We have, $\left(x^2+y^2\right)^2=x y$

On differentiating both sides w.r.t. $x$, we get

$$\frac{d}{d x}\left(x^2+y^2\right)^2=\frac{d}{d x}(x y)$$

$$ \begin{aligned} & \Rightarrow \quad 2\left(x^2+y^2\right) \cdot \frac{d}{d x}\left(x^2+y^2\right)=x \cdot \frac{d}{d x} y+y \cdot \frac{d}{d x} x \\ & \Rightarrow \quad 2\left(x^2+y^2\right) \cdot\left(2 x+2 y \frac{d y}{d x}\right)=x \frac{d y}{d x}+y \\ & \Rightarrow \quad 2 x^2 \cdot 2 x+2 x^2 \cdot 2 y \frac{d y}{d x}+2 y^2 \cdot 2 x+2 y^2 \cdot 2 y \frac{d y}{d x}=x \frac{d y}{d x}+y \\ & \Rightarrow \quad \frac{d y}{d x}\left[4 x^2 y+4 y^3-x\right]=y-4 x^3-4 x y^2 \\ & \therefore \quad \frac{d y}{d x}=\frac{\left(y-4 x^3-4 x y^2\right)}{\left(4 x^2 y+4 y^3-x\right)}\\ & \end{aligned}$$

We have, $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0\quad\text{.... (i)}$

On differentiating both sides w.r.t. $x$, we get

$$\frac{d}{d x}\left(a x^2\right)+\frac{d}{d x}(2 h x y)+\frac{d}{d x}\left(b y^2\right)+\frac{d}{d x}(2 g x)+\frac{d}{d x}(2 f y)+\frac{d}{d x}(c)=0$$

$$\begin{aligned} & \Rightarrow \quad 2 a x+2 h\left(x \cdot \frac{d y}{d x}+y \cdot 1\right)+b \cdot 2 y \frac{d y}{d x}+2 g+2 f \frac{d y}{d x}+0=0 \\ & \Rightarrow \quad \frac{d y}{d x}[2 h x+2 h y+2 f]=-2 a x-2 h y-2 g \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{-2(a x+h y+g)}{2(h x+b y+f)} \\ &=\frac{-(a x+h y+g)}{(h x+b y+f)}\quad\text{.... (ii)} \end{aligned}$$

$$\begin{aligned} &\text { Now, differentiating Eq. (i) w.r.t. y, we get }\\ &\begin{gathered} \quad \frac{d}{d y}\left(a x^2\right)+\frac{d}{d y}(2 h x y)+\frac{d}{d y}\left(b y^2\right)+\frac{d}{d y}(2 g x)+\frac{d}{d y}(2 f y)+\frac{d}{d y}(c)=0 \\ \Rightarrow \quad a \cdot 2 x \cdot \frac{d x}{d y}+2 h \cdot\left(x \cdot \frac{d}{d y} y+y \cdot \frac{d}{d y} x\right)+b \cdot 2 y+2 g \cdot \frac{d x}{d y}+2 f+0=0 \end{gathered} \end{aligned}$$

$$ \begin{aligned} & \Rightarrow \quad \frac{d x}{d y}[2 a x+2 h y+2 g]=-2 h x-2 b y-2 f \\ & \Rightarrow \quad \frac{d x}{d y}=\frac{-2(h x+b y+f)}{2(a x+h y+g)}=\frac{-(h x+b y+f)}{(a x+h y+g)} \quad\text{.... (iii)}\\ & \therefore \quad \frac{d y}{d x} \cdot \frac{d x}{d y}=\frac{-(a x+h y+g)}{(h x+b y+f)} \cdot \frac{-(h x+b y+f)}{(a x+h y+g)} \\ & =1=\mathrm{RHS}\quad\text{[using Eqs. (ii) and (iii)]} \end{aligned}$$

Hence proved.

$$\begin{aligned} &\text { We have, }\\ &\begin{aligned} x & =\mathrm{e}^{x / y} \\ \therefore\quad \frac{d}{d x} x & =\frac{d}{d x} e^{x / y} \end{aligned} \end{aligned}$$

$$\begin{array}{ll} \Rightarrow & 1=e^{x / y} \cdot \frac{d}{d x}(x / y) \\ \Rightarrow & 1=e^{x / y} \cdot\left[\frac{y \cdot 1-x \cdot d y / d x}{y^2}\right] \\ \Rightarrow & y^2=y \cdot e^{x / y}-x \cdot \frac{d y}{d x} \cdot e^{x / y} \end{array}$$

$$\begin{aligned} \Rightarrow \quad x \cdot \frac{d y}{d x} \cdot e^{x / y} & =y e^{x / y}-y^2 \\ \therefore \quad \frac{d y}{d x} & =\frac{y\left(e^{x / y}-y\right)}{x \cdot e^{x / y}} \\ & =\frac{\left(e^{x / y}-y\right)}{e^{x / y} \cdot \frac{x}{y}} \quad \left[\because x=\mathrm{e}^{x / y} \Rightarrow \log x=\frac{x}{y}\right]\\ & =\frac{x-y}{x \cdot \log x}\quad\text{Hence proved.} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text { We have, }\quad y^x & =\mathrm{e}^{y-x} \\ \Rightarrow\quad \log y^x & =\log e^{y-x} \\ \Rightarrow\quad x \log y & =y-x \cdot \log _e=(y-x) \quad \left[\because \log _e=1\right]\\ \Rightarrow\quad \log y & =\frac{(y-x)}{x}\quad\text{.... (i)} \end{aligned} \end{aligned}$$

Now, differentiating w.r.t. $x$, we get

$$\frac{d}{d y} \log y \cdot \frac{d y}{d x}=\frac{d}{d x} \frac{(y-x)}{x}$$

$$\begin{array}{ll} \Rightarrow & \frac{1}{y} \cdot \frac{d y}{d x}=\frac{x \cdot \frac{d}{d x}(y-x)-(y-x) \cdot \frac{d}{d x} \cdot x}{x^2} \\ \Rightarrow & \frac{1}{y} \frac{d y}{d x}=\frac{x\left(\frac{d y}{d x}-1\right)-(y-x)}{x^2} \end{array}$$

$$ \begin{array}{llrl} \Rightarrow & \frac{x^2}{y} \cdot \frac{d y}{d x} =x \frac{d y}{d x}-x-y+x \\ \Rightarrow & \frac{d y}{d x}\left(\frac{x^2}{y}-x\right) =-y \end{array}$$

$\therefore \quad \frac{d y}{d x}=\frac{-y^2}{x^2-x y}=\frac{-y^2}{x(x-y)}$

$$\begin{aligned} &\begin{aligned} & =\frac{y^2}{x(y-x)} \cdot \frac{x}{x}=\frac{y^2}{x^2} \cdot \frac{1}{\frac{(y-x)}{x}} \\ & =\frac{(1+\log y)^2}{\log y}\left[\because \log y=\frac{y-x}{x} \log y=\frac{y}{x}-1 \Rightarrow 1+\log y=\frac{y}{x}\right] \end{aligned}\\ &\text { Hence proved. } \end{aligned}$$

We have, $y=(\cos x)^{(\cos x)^{\left.(\cos x)^{-\infty}\right)}}$

$$\begin{array}{ll} \Rightarrow & y=(\cos x)^y \\ \therefore & \log y=\log (\cos x)^y \\ \Rightarrow & \log y=y \log \cos x \end{array}$$

$$\begin{aligned} &\text { On differentiating w.r.t. } x \text {, we get }\\ &\begin{aligned} & \frac{1}{y} \cdot \frac{d y}{d x}=y \cdot \frac{d}{d x} \log \cos x+\log \cos x \cdot \frac{d y}{d x} \\ \Rightarrow \quad & \frac{1}{y} \cdot \frac{d y}{d x}=\frac{y}{\cos x} \cdot \frac{d}{d x} \cos x+\log \cos x \cdot \frac{d y}{d x} \end{aligned} \end{aligned}$$

$$\begin{aligned} \Rightarrow \quad \frac{d y}{d x}\left[\frac{1}{y}-\log \cos x\right] & =\frac{-y \sin x}{\cos x}=-y \tan x \\ \therefore \quad \frac{d y}{d x} & =\frac{-y^2 \tan x}{(1-y \log \cos x)} \\ & =\frac{y^2 \tan x}{y \log \cos x-1}\quad \text{Hence proved.} \end{aligned}$$