We have, $\sec (x+y)=x y$

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

$$\frac{d}{d x} \sec (x+y)=\frac{d}{d x}(x y)$$

$$\begin{array}{ll} \Rightarrow & \sec (x+y) \cdot \tan (x+y) \cdot \frac{d}{d x}(x+y)=x \cdot \frac{d}{d x} y+y \cdot \frac{d}{d x} x \\ \Rightarrow & \sec (x+y) \cdot \tan (x+y) \cdot\left(1+\frac{d y}{d x}\right)=x \frac{d y}{d x}+y \end{array}$$

$$\begin{array}{ll} \Rightarrow & \sec (x+y) \tan (x+y)+\sec (x+y) \cdot \tan (x+y) \cdot \frac{d y}{d x}=x \frac{d y}{d x}+y \\ \Rightarrow & \frac{d y}{d x}[\sec (x+y) \cdot \tan (x+y)-x]=y-\sec (x+y) \cdot \tan (x+y) \\ \therefore & \frac{d y}{d x}=\frac{y-\sec (x+y) \cdot \tan (x+y)}{\sec (x+y) \cdot \tan (x+y)-x} \end{array}$$

We have, $\tan ^{-1}\left(x^2+y^2\right)=a$

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

$$\frac{d}{d x} \tan ^{-1}\left(x^2+y^2\right)=\frac{d}{d x}(a)$$

$$\begin{array}{rr} & \frac{d}{d x} \tan ^{-1}\left(x^2+y^2\right)=\frac{d}{d x}(a) \\ \Rightarrow & \frac{1}{1+\left(x^2+y^2\right)^2} \cdot \frac{d}{d x}\left(x^2+y^2\right)=0 \\ \Rightarrow & 2 x+\frac{d}{d y} y^2 \cdot \frac{d y}{d x}=0 \\ \therefore & 2 y \cdot \frac{d y}{d x}=-2 x \\ \Rightarrow & \frac{d y}{d x}=\frac{-2 x}{2 y}=\frac{-x}{y} \end{array}$$

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.

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

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

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