We have,
$$
\begin{array}{rlrl}
& \bar{z}^{\frac{1}{3}} & =a+i b \\
\Rightarrow & \quad \bar{z} & =(a+i b)^3 \\
\Rightarrow & x-i y & =(a+i b)^3 \\
\Rightarrow & x-i y & =a^3+i b^3+3 a^2(i b)+3 a\left(i^2 b^2\right) \\
\Rightarrow & x-i y & =a^3-i b^3+3 a^2 b i-3 a b^2 \\
\Rightarrow & x-i y & =\left(a^3-3 a b^2\right)+i\left(3 a^2 b-b^3\right) \\
\Rightarrow & & x & =a^3-3 a b^2 \text { and } y=-3 a^2 b+b^3
\end{array}
$$
$$
\Rightarrow \quad \frac{x}{a}=a^2-3 b^2 \text { and } \frac{y}{b}=-3 a^2+b^2
$$
Now, $\frac{x}{a}+\frac{y}{b}=a^2-3 b^2-3 a^2+b^2$
$$
\begin{aligned}
& =-2 a^2-2 b^2 \\
& \Rightarrow \quad \frac{x}{a}+\frac{y}{b}=-2\left(a^2+b^2\right) \\
& \therefore \quad \frac{1}{a^2+b^2}\left(\frac{x}{a}+\frac{y}{b}\right)=-2 \\
&
\end{aligned}
$$