Answer
False
Work Step by Step
Let $\left(x^{2}+y^{2}-\frac{1}{4}\right)^{2}= f(x, y)$
\[
f_{x}(x, y)=4 x\left(x^{2}+y^{2}-\frac{1}{4}\right) \quad \text { and } \quad f_{y}(x, y)=4 y\left(x^{2}+y^{2}-\frac{1}{4}\right)
\]
By theorem $13.8 .5,$ a point $\left(x_{0}, y_{0}\right)$ in the domain of a function $f(x, y)$ is called a critical point of the function if $f_{x}\left(x_{0}, y_{0}\right)=0$ and $f_{y}\left(x_{0}, y_{0}\right)=0$ or if one or both partial derivatives do not exist at $\left(x_{0}, y_{0}\right)$
So here $f_{x}(x, y)=0=4 x\left(x^{2}+y^{2}-\frac{1}{4}\right) \quad$ and $\quad f_{y}(x, y)=0=4 y\left(x^{2}+y^{2}-\frac{1}{4}\right)$
\[
0=x^{2}+y^{2}-\frac{1}{4} \quad \text { and }0= \quad x^{2}+y^{2}-\frac{1}{4}
\]
so every point on $\frac{1}{4}=x^{2}+y^{2}$ is a critical point of $f$
