Given,
$$
\begin{aligned}
z & =x-i y \text { and } z^{\frac{1}{3}}=a+i b \\
z^{1 / 3} & =a+i b
\end{aligned}
$$
Take cube on both sides, we get
$$
\begin{array}{ll}
\left(z^{1 / 3}\right)^3=(a+i b)^3 & \\
\Rightarrow z=a^3+(i b)^3+3 a^2 i b+3 a(i b)^2 & \\
\Rightarrow z=a^3+i^3 b^3+3 a^2 i b+3 a b^2 i^2 & \\
\Rightarrow z=a^3-i b^3+3 a^2 b i-3 a b^2 & {\left[\because i^2=-1\right]} \\
\Rightarrow z=\left(a^3-3 a b^2\right)-i\left(b^3-3 a^2 b\right) & \\
\Rightarrow x-i y=\left(a^3-3 a b^2\right)-i\left(b^3-3 a^2 b\right) \quad[\because z=x-i y]
\end{array}
$$
Here, $x=a^3-3 a b^2$ and $y=b^3-3 a^2 b$
Now,
$$
\begin{aligned}
\frac{\left(\frac{x}{a}+\frac{y}{b}\right)}{a^2+b^2} & =\frac{\left(\frac{a^3-3 a b^2}{a}+\frac{b^3-3 a^2 b}{b}\right)}{a^2+b^2} \\
& =\frac{a^2-3 b^2+b^2-3 a^2}{a^2+b^2} \\
& =\frac{-2 a^2-2 b^2}{a^2+b^2}=\frac{-2\left(a^2+b^2\right)}{a^2+b^2}=-2
\end{aligned}
$$