Answer
The equation for the tangent line is $y=x.$
Work Step by Step
The equation for the tangent line at $x_0$ is:
$$y-f(x_0)=f'(x_0)(x-x_0)$$
Here is $x_0=0$ and $f(x)=\frac{x}{\sqrt{1-x^2}}.$
$$f'(x)=\Big(\frac{x}{\sqrt{1-x^2}}\Big)'=\frac{x'\sqrt{1-x^2}-x(\sqrt{1-x^2})'}{(\sqrt{1-x^2})^2}=\frac{\sqrt{1-x^2}-x\frac{1}{2\sqrt{1-x^2}}(1-x^2)'}{1-x^2}=\frac{1-x^2-x\cdot\frac{(-2x)}{2}}{(1-x^2)^{3/2}}=\frac{1}{(1-x^2)^{3/2}}$$
Now, for $x=0$ we have:
$$f(0)=\frac{0}{\sqrt{1-0^2}}=\frac{0}{1}=0$$
$$f'(0)=\frac{1}{(1-0^2)^{3/2}}=\frac{1}{1}=1$$
So, the equation for the tangent line is:
$$y-0=1\cdot(x-0)\Rightarrow y=x$$
