Sketching x√(1 − x²)

Domain and parity. $f$ is defined on $[-1,1]$ and is odd.

Derivative. On $(-1,1)$,
$$f'(x)=\sqrt{1-x^2}-\frac{x^2}{\sqrt{1-x^2}}=\frac{1-2x^2}{\sqrt{1-x^2}}.$$
Positive for $|x|<\frac1{\sqrt2}$ and negative for $|x|>\frac1{\sqrt2}$. Maximum $f\left(\frac1{\sqrt2}\right)=\frac12$, minimum $f\left(-\frac1{\sqrt2}\right)=-\frac12$.

Endpoints. $f(1)=0$ and $\dfrac{f(x)-f(1)}{x-1}=-x\sqrt{\dfrac{1+x}{1-x}}\to-\infty$ as $x\to1^-$: the tangent at $(1,0)$ is vertical. Likewise at $(-1,0)$.

Convexity. $f''(x)=\dfrac{x(2x^2-3)}{(1-x^2)^{3/2}}$, and $2x^2-3<0$ on the domain: concave on $[0,1)$, convex on $(-1,0]$, inflection point at the origin with slope $f'(0)=1$.

Graph. An "S" shape from $(-1,0)$ down to the minimum $-\frac12$, up through the origin to the maximum $\frac12$, then down to $(1,0)$, with vertical tangents at both endpoints.

Remark. The union of this graph and its reflection $y=-x\sqrt{1-x^2}$ is the figure eight (closed curve) $y^2=x^2-x^4$.