Below is the surface graph of the function

$$f(x,y)=\tfrac{1}{2}\bigl(1-(x-\tfrac{1}{2})^2-(y-\tfrac{1}{2})^2\bigr)$$

This function has first partial derivatives:

$$f_x(x,y)=\tfrac{1}{2}-x\qquad\text{and}\qquad f_y(x,y)=\tfrac{1}{2}-y$$ Both derivatives are zero at the point $(\tfrac{1}{2},\tfrac{1}{2})$ and thus is the tangent plane at the point $(\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2})$ horizontal. The point $(\tfrac{1}{2},\tfrac{1}{2})$ is a critical point and the function has a maximum value in that point.

Below is the surface graph of the function

$$f(x,y)=\tfrac{1}{2}\bigl(1-(x-\tfrac{1}{2})^2+(y-\tfrac{1}{2})^2\bigr)$$

This function has first partial derivatives:

$$f_x(x,y)=\tfrac{1}{2}-x\qquad\text{and}\qquad f_y(x,y)=y-\tfrac{1}{2}$$ Both derivatives are zero at the point $(\tfrac{1}{2},\tfrac{1}{2})$ and thus is the tangent plane at the point $(\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2})$ horizontal. But the level curve through this point in the $x$ direction is a mountain parabola, and the the level curve in the $y$direction is a valley parabola. Such a point is called a saddle point.

Up to now, critical points were always one or more disjoint points. But this need not always be the case. Consider, for example, the function

$$f(x,y)=\tfrac{1}{2}\bigl(1-\sin(2x^2-y-1)\bigr)$$ This function has first partial derivatives: $$f_x(x,y)=-2x\cos(2x^2-y-1)\qquad\text{andn}\qquad f_y(x,y)=\tfrac{1}{2}\cos(2x^2-y-1)$$ The gradient is the zero vector in all points $(x,y)$ for which the following is true: $$\cos(2x^2-y-1)=0$$ The critical points in this case are not isolated points, but points on curves, namely the parabolas $$2x^2-y-1=\tfrac{1}{2}\pi +k\pi\qquad\text{with integer }k$$ One of these parabolas is $$y=2x^2-1-\tfrac{1}{2}\pi$$ is drawn as a thick line in the graph below; it corresponds with the level curve of height $1$. In all points of this parabola, the function reaches a maximum value. The corresponding tangent plane is the plane $z=1$, the top face of the cube which is drawn around it.