if:
f(sinx)=(sinx)²
calculate : f"(x)
No! That would be correct if $\displaystyle f(x)= sin^2(x)$ but the original post said $\displaystyle f(sin(x))= sin^2(x)$. As Idea said, that implies that $\displaystyle f(x)= x^2$ so $\displaystyle f'(x)= 2x$ and $\displaystyle f''(x)= 2$.
and
$\displaystyle \dfrac{df}{dx} = \dfrac{df}{dy} \dfrac{dy}{dx} = 2y \cdot cos(x) = 2~sin(x)~cos(x) = sin(2x)$
Can you finish?
-Dan