If $x^p{y}^q={(x+y)}^{p+q},$ then $\frac{dy}{dx}=$

Taking $log$ both sides, $plogx+qlogy=(p+q)log(x+y)$

$$\begin{gathered} \Rightarrow \frac{p}{x}+\frac{q}{y}\frac{dy}{dx}=\frac{p+q}{x+y}(1+\frac{dy}{dx}) \\ \Rightarrow \left(\frac{p}{x}-\frac{p+q}{x+y}\right)=\left(\frac{p+q}{x+y}-\frac{q}{y}\right)\frac{dy}{dx} \\ \begin{array}{l}\Rightarrow \frac{px+py-px-qx}{x(x+y)}=\frac{py+qy-qx-qy}{y(x+y)}\frac{dy}{dx}\\ \Rightarrow \frac{py-qx}{x(x+y)}=\frac{py-qx}{y(x+y)}\frac{dy}{dx}\end{array} \\ \Rightarrow \frac{{\displaystyle dy}}{{\displaystyle dx}}=\frac{{\displaystyle y}}{{\displaystyle x}} \end{gathered}$$