Answer
$$-\sin[g(0)]$$
Work Step by Step
Given $$ g(x)+x\sin g(x)=x^2$$
differentiate the both sides
\begin{align*}
g'(x)+xg'(x) \cos [g(x)]+\sin[ g(x)]&=2x\\
g'(x)[x \cos [g(x)] +1]&=2x-\sin[g(x)]\\
g'(x)&=\frac{2x-\sin[g(x)]}{x \cos [g(x)] +1}
\end{align*}
then $$g'(0)=\frac{2(0)-\sin[g(0)]}{(0) \cos [g(0)] +1}= -\sin[g(0)]$$
