Answer
Using the First Derivative Test or the Second Derivative Test, we can determine that $f$ has a local maximum at $x = 0$ and $f$ has a local minimum at $x = 2$
Work Step by Step
$f(x) = \frac{x^2}{x-1}$
First Derivative Test:
$f'(x) = \frac{(2x)(x-1)-(x^2)(1)}{(x-1)^2}$
$f'(x) = \frac{x^2-2x}{(x-1)^2} = 0$
$x^2-2x = 0$
$x(x-2) = 0$
$x = 0, 2$
As $x \to 0^-,$ then $f'(x) \gt 0$
As $x \to 0^+,$ then $f'(x) \lt 0$
$f$ has a local maximum at $x = 0$
As $x \to 2^-,$ then $f'(x) \lt 0$
As $x \to 2^+,$ then $f'(x) \gt 0$
$f$ has a local minimum at $x = 2$
Second Derivative Test:
$f'(0) = 0$
$f'(2) = 0$
$f''(x) = \frac{(2x-2)(x-1)^2-(x^2-2x)(2)(x-1)}{(x-1)^4}$
$f''(x) = \frac{(2x-2)(x-1)-(x^2-2x)(2)}{(x-1)^3}$
$f''(x) = \frac{2x^2-4x+2-2x^2+4x}{(x-1)^3}$
$f''(x) = \frac{2}{(x-1)^3}$
At $x = 0,~~f''(x) \lt 0$
$f$ has a local maximum at $x = 0$
At $x = 2,~~f''(x) \gt 0$
$f$ has a local minimum at $x = 2$