Find the critical points of $f(x)=-6x^4+2x^2-8$ and apply the Second Derivative Test (if possible) to determine whether each of them corresponds to a local minimum or maximum.
(Use symbolic notation and fractions where needed. Give your answer in the form of a comma separated list. Enter DNE if there are no critical points.)
the critical points with a local minimum at $x=$
the critical points with a local maximum at $x=$

Step 1 ：Given the function $f(x)=-6x^4+2x^2-8$, we first need to find the derivative of the function.

Step 2 ：The derivative of the function $f^{\prime }(x)=-24x^3+4x$.

Step 3 ：To find the critical points, we set the derivative equal to zero and solve for $x$. The solutions are the critical points: $x=0,-\sqrt{6}/6,\sqrt{6}/6$.

Step 4 ：We then find the second derivative of the function $f^{″}(x)=4-72x^2$.

Step 5 ：Applying the Second Derivative Test, we find that the second derivative at $x=0$ is positive, so the function has a local minimum at that point.

Step 6 ：The second derivative at $x=-\sqrt{6}/6,\sqrt{6}/6$ is negative, so the function has a local maximum at those points.

Step 7 ：$\boxed{{\displaystyle \text{The critical points with a local minimum at }x=0\text{ and the critical points with a local maximum at }x=-\frac{\sqrt{6}}{6},\frac{\sqrt{6}}{6}}}$