Here, $\mathrm{y}^{\mathrm{x}}=\mathrm{e}^{\mathrm{y}-\mathrm{x}}$
Taking log on both sides, we get
$\begin{aligned}
& \log y^x=\log e^{y-x} \\
& \quad\left(\because \log a^b=b \log a \text { and } \log e=1\right)
\end{aligned}$
$\Rightarrow x \log y=(y-x) \log e \Rightarrow x \log y=y-x \ldots(i)$
On differentiating w.r.t. $x$, we get
$\frac{d}{d x}(x \log y)=\frac{d}{d x}(y-x)$
(using product rule)
$\begin{aligned}
& \Rightarrow x\left(\frac{1}{y}\right) \frac{d y}{d x}+\log y(1)=\frac{d y}{d x}-1 \\
& \Rightarrow \frac{d y}{d x}\left(\frac{x}{y}-1\right)=-1-\log y \\
& \Rightarrow \frac{d y}{d x}\left[\frac{y}{(1+\log y) y}-1\right]=-(1+\log y) \\
& \qquad \quad\left[\because \text { from eq. }(i) \cdot x=\frac{y}{(1+\log y)}\right] \\
& \Rightarrow \frac{d y}{d x}\left[\frac{1-1-\log y}{1+\log y}\right]=-(1+\log y)
\end{aligned}$
$\Rightarrow \frac{d y}{d x}=-\frac{(1+\log y)^2}{-\log y}$
$\Rightarrow \frac{d y}{d x}=\frac{(1+\log y)^2}{\log y}$