Answer
\[
\begin{array}{l}
-\sin (x+y)^{3}+\sin (x-y)^{3}= f_{x}(x, y)\\
-\left(\sin (-y+x)^{3}+\sin (y+x)^{3}\right)=f_{y}(x, y)
\end{array}
\]
Work Step by Step
We are given that
\[
\int_{x+y}^{x-y} \sin t^{3} d t=f(x, y)
\]
We know that
\[
f_{x}(x, y)=\frac{\partial}{\partial x}\left(\int_{x+y}^{x-y} \sin t^{3} d t\right)
\]
\[
\int_{h(x)}^{g(x)} f(t) d t=f(x)
\]
and then
\[
g^{\prime}(x) \int(g(x))-h^{\prime}(x) f(h(x))=f^{\prime}(x)
\]
Thus
\[
\begin{array}{l}
\frac{d}{d x}(x-y) * \sin (x-y)^{3}-\frac{d}{d x}(x+y) * \sin (x+y)^{3}=f_{x}(x, y) \\
-\sin (x+y)^{3}+\sin (x-y)^{3}=f_{x}(x, y)
\end{array}
\]
We obtain $f_{y}(x, y)$
\[
f_{y}(x, y)=\frac{d}{d y}(-y+x) * \sin (-y+x)^{3}-\frac{d}{d y}(y+x) * \sin (y+x)^{3}
\]
\[
-\left(\sin (-y+x)^{3}+\sin (y+x)^{3}\right)=f_{y}(x, y)
\]
