From $f_x(x,y,z)=0$, $f_y(x,y,z)=0$ and $f_z(x,y,z)=0$ we find $(\frac12, 1, 1)$ and $(-\frac12, -1, -1)$. Since for $x,y,z>0$ $$f(x,y,z)=x+\frac{y^2}{4x}+\frac{z^2}{2y}+\frac{z^2}{2y}+\frac{1}{2z}+\frac{1}{2z}+\frac{1}{2z}+\frac{1}{2z} \ge 8 \sqrt[8]{\frac{1}{2^8}}=4=f\left(\frac12, 1, 1\right)$$ the function has a local minimum at $(\frac12, 1, 1)$. Similarly $(-\frac12, -1, -1)$ is a point of local maximum.