If $u=\frac{x}{y^2}-\frac{y}{x^2}$ then verify $\frac{{\partial }^{2}u}{\partial x\partial y}=\frac{{\partial }^{2}u}{\partial y\partial x}$

Given $u=\frac{x}{y^2}-\frac{y}{x^2}$

We have to verify $\frac{{\partial }^{2}u}{\partial x\partial y}=\frac{{\partial }^{2}u}{\partial y\partial x}$

Consider $\frac{{\partial }^{2}u}{\partial x\partial y}$

$\frac{{\partial }^{2}u}{\partial x\partial y}=\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial y}\right)$

$\frac{\partial u}{\partial y}=-\frac{2x}{y^3}-\frac{1}{x^2}$

$\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial y}\right)=\frac{\partial }{\partial x}(-\frac{2x}{y^3}-\frac{1}{x^2})$

$=-\frac{2}{y^3}+\frac{2}{x^3}$

Thus $\frac{{\partial }^{2}u}{\partial x\partial y}=-\frac{2}{y^3}+\frac{2}{x^3}...\left(1\right)$

Consider $\frac{{\partial }^{2}u}{\partial y\partial x}$

$\frac{{\partial }^{2}u}{\partial y\partial x}=\frac{\partial }{\partial y}\left(\frac{\partial u}{\partial x}\right)$

$\frac{\partial u}{\partial x}=\frac{1}{y^2}+\frac{2y}{x^3}$

$\frac{\partial }{\partial y}\left(\frac{\partial u}{\partial x}\right)=\frac{\partial }{\partial y}(-\frac{1}{y^2}+\frac{2y}{x^3})$

$=-\frac{2}{y^3}+\frac{2}{x^3}$

Thus $\frac{{\partial }^{2}u}{\partial y\partial x}=-\frac{2}{y^3}+\frac{2}{x^3}...\left(2\right)$

From $\left(1\right),\left(2\right)$ we get

$\frac{{\partial }^{2}u}{\partial x\partial y}=\frac{{\partial }^{2}u}{\partial y\partial x}$

Hence verified.