Answer
$$y=-\sqrt{-2 x \cos x+2 \sin x+1} $$
Work Step by Step
Given $$\frac{d y}{d x}=\frac{x \sin x}{y}, \quad y(0)=-1$$
Separating the variables, we get
\begin{align*}
\int ydy &= \int x\sin xdx\\
\frac{1}{2} y^2&= \int x\sin xdx
\end{align*}
Use integration by parts
\begin{align*}
u&= x\ \ \ \ \ \ dv=\sin xdx\\
du&=dx\ \ \ \ \ v=-\cos x
\end{align*}
Then
\begin{align*}
\int x\sin xdx&= -x\cos x +\int \cos xdx\\
&= -x\cos x +\sin x+C
\end{align*}
Hence
\begin{align*}
\frac{1}{2} y^2&= -x\cos x +\sin x+C
\end{align*}
At $x=0$, $y=-1$, then $C=\dfrac{1}{2} $ and
$$y=- \sqrt{-2 x \cos x+2 \sin x+1} $$
