[tex]y=(4x^2-1)^4\\
y'=4(4x^2-1)^3\cdot8x=32x(4x^2-1)^3\\\\
32x(4x^2-1)^3=0\\
x=0 \vee 4x^2-1=0\\
4x^2-1=0\\
4x^2=1\\
x^2=\dfrac{1}{4}\\
x=-\dfrac{1}{2} \vee x=\dfrac{1}{2}\\
x\in\left\{-\dfrac{1}{2},0,\dfrac{1}{2}\right\}\leftarrow \text{stationary points}\\\\
[/tex]
[tex]y''=32((4x^2-1)^3+x\cdot3(4x^2-1)^2\cdot8x)\\
y''=32(4x^2-1)^2(4x^2-1+24x^2)\\
y''=32(4x^2-1)^2(28x^2-1)\\\\
y''(0)=32(4\cdot0^2-1)^2(28\cdot0^2-1)=32\cdot(-1)^2\cdot(-1)=-32[/tex]
[tex]-32<0 \Rightarrow[/tex] there's a (local) maximum at [tex]x=0[/tex].
For [tex]x\in\left\{-\dfrac{1}{2},\dfrac{1}{2}\right\}[/tex], [tex]y''=0[/tex] because the factor [tex](4x^2-1)^2=0[/tex].
You have to check values of [tex]y'[/tex] around [tex]x\in\left\{-\dfrac{1}{2},\dfrac{1}{2}\right\}[/tex].
[tex]\forall x\in(\infty,-\frac{1}{2})\, y'<0 \Rightarrow y\searrow\\
\forall x\in(-\frac{1}{2},0)\, y'>0 \Rightarrow y\nearrow\\\\
\forall x\in(0,\frac{1}{2})\, y'<0 \Rightarrow y\searrow\\
\forall x\in(\frac{1}{2},\infty)\, y'>0 \Rightarrow y\nearrow
[/tex]
The above means that at [tex]x=-\dfrac{1}{2}[/tex] and [tex]x=-\dfrac{1}{2}[/tex] there are (global) maxima.
