$(x+i y)^{3}=u+i v$

$\Rightarrow x^{3}+(i y)^{3}+3 \cdot x \cdot i y(x+i y)=u+i v$

$\Rightarrow x^{3}+i^{3} y^{3}+3 x^{2} y i+3 x y^{2} i^{2}=u+i v$

$\Rightarrow x^{3}-i y^{3}+3 x^{2} y i-3 x y^{2}=u+i v$

$\Rightarrow\left(x^{3}-3 x y^{2}\right)+i\left(3 x^{2} y-y^{3}\right)=u+i v$

On equating real and imaginary parts, we obtain

$u=x^{3}-3 x y^{2}, v=3 x^{2} y-y^{3}$

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

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

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

$=4 x^{2}-4 y^{2}$

$=4\left(x^{2}-y^{2}\right)$

$\therefore \frac{u}{x}+\frac{v}{y}=4\left(x^{2}-y^{2}\right)$

Hence, proved.