Answer
Local minima: $(-1,-1)$ and $(1,-1)$ (both are absolute minima)
Local maximum: $(0,0)$
Inflection points: $( \displaystyle \pm\frac{1}{\sqrt{3}},\ -\displaystyle \frac{5}{9})$
Work Step by Step
$y=f(x)=x^{4}-2x^{2}$
$f'(x)=4x^{3}-4x=4x(x^{2}-1)$
$f''(x)=12x^{2}-4=4(3x^{2}-1)$
Analyzing $f'(x)$, we see that it is defined everywhere,
and has zeros $x=-1,0$ and $ 1\qquad$... (critical points)
Applying the 2nd derivative test,
$\left\{\begin{array}{lllll}
x=-1, & f''(-1)=8\gt 0 & \Rightarrow(\min), & f(-1)=-1 & (-1,-1)\\
x=0, & f''(0)=-4 \lt 0 & \Rightarrow(\max), & f(0)=0 & (0,0)\\
x=1, & f''(1)=8\gt 0 & \Rightarrow(\min), & f(1)=-1 & (1,-1)
\end{array}\right.$
$f: \left[\begin{array}{cccccccc}
& & -- & | & ++ & | & -- & | & ++ & \\
& (-\infty & \searrow & -1 & \nearrow & 0 & \searrow & 1 & \nearrow & \infty)
\end{array}\right]$
$f''(x)=0$ for $ x= \displaystyle \pm\frac{1}{\sqrt{3}} \qquad f( \pm\frac{1}{\sqrt{3}} )=-\frac{5}{9},\quad$
$(\displaystyle \pm\frac{1}{\sqrt{3}},-\frac{5}{9})$ are points of inflection.
For graphing purposes:
Concavity and points of inflection:
$\left[\begin{array}{ccccccc}
f''(x): & -\infty & ++ & \frac{1}{\sqrt{3}} & -- & \frac{1}{\sqrt{3}} & ++ & \infty\\
f(x): & & \cup & inf. & \cap & inf. & \cup &
\end{array}\right]$
With these data, we graph $f(x).$
