$$f(x,y)=\frac{2}{x^2+(y-1)^2+1}-\frac{1}{x^2+(y+1)^2+1} \tag{1}\label{1}$$
will do. 

Indeed, this function $f$ has exactly three critical points: a saddle point $(0,\approx-8.54)$, a point of a local minimum at $(0,\approx-1.14)$, and a point of a local maximum at $(0,\approx1.04)$. 

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Below are the calculations, in Mathematica:

[![enter image description here][1]][1]
[![enter image description here][2]][2]

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To have exactly three critical points (again, a saddle, a local minimum, and a local maximum) which are noncollinear, one can use 
$$f(x,y)=\frac{2}{x^2/2+x/10+(y-1)^2+1}-\frac{1}{x^2+(y+1)^2+1}$$
instead. Then, with a suitable affine transformation $A$, one can have any given three noncollinear points on the plane to be the three critical points (again, a saddle, a local minimum, and a local maximum) of the function $\mathbb R^2\ni z\mapsto f(Az)$. 

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To have exactly three critical points (again, a saddle, a local minimum, and a local maximum) which are in arbitrary collinear positions -- except that the saddle point is not between the two extrema points, one can use the function $f$ defined by \eqref{1} together with a simple diffeomorphism of $\mathbb R$. 

It remains unclear if we can have exactly three collinear critical points (again, a saddle, a local minimum, and a local maximum) such that the saddle point is between the two extrema points. 

  [1]: https://i.sstatic.net/LAsPA.png
  [2]: https://i.sstatic.net/P8cgD.png