Answer
$$f^{-1}(x)=-\sqrt{x}$$
Work Step by Step
$$y=f(x)=x^2\hspace{1cm}x\le0$$
To find its inverse:
1) Solve for $x$ in terms of $y$:
$$y = x^2$$
Since our domain for $x$ is $(-\infty,0]$, we take the negative value of $x$ here, which means
$$x=-\sqrt{y}$$
2) Interchange $x$ and $y$:
$$y=-\sqrt{x}$$
Therefore, the inverse of function $y=f(x)=x^2$, $x\le0$ is the function $y=-\sqrt{x}$. In other words, $$f^{-1}(x)=-\sqrt{x}$$
