$\begin{aligned}
&(x+i y)^3=x^3+3 x^2 i y+3 x \cdot(i y)^2+(i y)^3 \\
&\quad=x^3+3 x^2 i y+3 x i^2 y^2+i^3 y^3 \\
&\quad=x^3+i 3 x^2 y-3 x y^2-i y^3 \\
&\quad=\left(x^3-3 x y^2\right)+\left(3 x^2 y-y^3\right) i=u+i v
\end{aligned}$
Equating real and imaginary parts
$u=x^3-3 x y^2$ or $\frac{u}{x}=x^2-3 y^2 \quad \ldots (i)$
and $v=3 x^2 y-y^3$ or $\frac{v}{y}=3 x^2-y^2 \quad \ldots (ii)$
Adding equations (i) \& (ii)
$\begin{aligned}
\frac{u}{x}+\frac{v}{y} &=x^2-3 y^2+3 x^2-y^2 \\
\Rightarrow \frac{u}{x}+\frac{v}{y} &=4\left(x^2-y^2\right) \quad \text { Hence proved }
\end{aligned}$