We have,
$$
\begin{array}{r}
\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=\sqrt{a} \\
\left(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}\right)^{2}=a \\
\frac{x}{y}+\frac{y}{x}+2=a
\end{array}
$$
Differentiating both sides w.r.t. $x$,
$\frac{y-x \frac{d y}{d x}}{y^{2}}+\frac{x \frac{d y}{d x}-y}{x^{2}}=0$
$\Rightarrow \quad \frac{y-x \frac{d y}{d x}}{y^{2}}=-\frac{\left(x \frac{d y}{d x}-y\right)}{x^{2}}$
$\Rightarrow \quad x y^{2} \frac{d y}{d x}-y^{3}=-x^{2} y+x^{3} \frac{d y}{d x}$
$\Rightarrow \quad \frac{d y}{d x}\left(x y^{2}-x^{3}\right)=-x^{2} y+y^{3}$
$\Rightarrow \quad \frac{d y}{d x}=\frac{y\left(y^{2}-x^{2}\right)}{x\left(y^{2}-x^{2}\right)}$
$\Rightarrow \quad \frac{d y}{d x}=\frac{y}{x}$
Now, $y \cdot \frac{d x}{d y}=y \cdot \frac{x}{y}=x$