Given that, \( x^{m} y^{n}=(x+y)^{m+n} \)
Taking log both the sides, we have
\( m \log x+n \log y=(m+n) \log (x+y) \)
Differentiating both the sides with respect to \( x \), we get
\[
\begin{array}{l}
\frac{m}{x}+\left(\frac{n}{y}\right) \frac{d y}{d x}=(m+n)\left(\frac{1+\frac{d y}{d x}}{x+y}\right) \\
\Rightarrow \frac{m}{x}-\frac{m+n}{x+y}=\left(\frac{m+n}{x+y}\right) \frac{d y}{d x}-\left(\frac{n}{y}\right) \frac{d y}{d x} \\
\Rightarrow \frac{m x+m y-m x-n x}{x(x+y)}=\left(\frac{m y+n y-n x-n y}{(x+y) y}\right) \frac{d y}{d x} \\
\Rightarrow \frac{1}{x}=\frac{1}{y} \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{y}{x}
\end{array}
\]