It is given that

$x=\frac{{\sin }^3\theta }{{\cos }^2\theta }$

$y=\frac{{\cos }^3\theta }{{\sin }^2\theta }$

The summation of$x$ and$y$ is

$x+y=\frac{{\sin }^3\theta }{{\cos }^2\theta }+\frac{{\cos }^3\theta }{{\sin }^2\theta }$

$=\frac{\sin \theta \left(1-{\cos }^2\theta \right)}{{\cos }^2\theta }+\frac{\cos \theta \left(1-{\sin }^2\theta \right)}{{\sin }^2\theta }$

$=\sin \theta \left(\frac{1}{{\cos }^2\theta }-1\right)+\cos \theta \left(\frac{1}{{\sin }^2\theta }-1\right)$

$=\frac{\sin \theta }{{\cos }^2\theta }+\frac{\cos \theta }{{\sin }^2\theta }-(\sin \theta +\cos \theta )$

$=\frac{\sin \theta }{{\cos }^2\theta }+\frac{\cos \theta }{{\sin }^2\theta }-\frac{1}{2}$

$=\frac{{\sin }^3\theta +{\cos }^3\theta }{{\cos }^2\theta {\sin }^2\theta }-\frac{1}{2}$

$\frac{=(\sin \theta +\cos \theta )\left({\sin }^2\theta +{\cos }^2\theta -\sin \theta \cos \theta \right)}{{\sin }^2\theta {\cos }^2\theta }-\frac{1}{2}$

$=\frac{\frac{1}{2}(1-\sin \theta \cos \theta )}{(\sin \theta \cos \theta {)}^2}-\frac{1}{2} ...\left(\mathrm{i}\right)$                   

It is given that

$\sin \theta +\cos \theta =\frac{1}{2}$

Squaring both sides

${\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta =\frac{1}{4}$

$\Rightarrow 1+2\sin \theta \cos \theta =\frac{1}{4}$

$\Rightarrow \sin \theta \cos \theta =\frac{-3}{8}$

Substitute the values in Equation $\left(\mathrm{i}\right)$

$x+y=\frac{\frac{1}{2}\left(1-\left(\frac{-3}{8}\right)\right)}{{\left(\frac{-3}{8}\right)}^2}-\frac{1}{2}$

$\Rightarrow x+y=\frac{79}{18}$