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$$f(x,y)=\frac{2}{x^2+(y-1)^2+1}-\frac{1}{x^2+(y+1)^2+1} \tag{1}\label{1}$$ 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 $(0,\approx1.04)$.

For any given three distinct points on the plane to be exactly the critical points (again, a saddle, a local minimum, and a local maximum) of a function, use $f\circ g$ instead of $f$, where $g$ is a diffeomorphism of the plane moving the given three distinct points to the critical points of $f$.

Below are the calculations for $f$, in Mathematica:

$\newcommand\R{\mathbb R}$ $$f(x,y)=\frac{2}{x^2+(y-1)^2+1}-\frac{1}{x^2+(y+1)^2+1} \tag{1}\label{1}$$ 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 $(0,\approx1.04)$.

For any given three distinct points $P,Q,R$ in $\R^2$ to be exactly the critical points of a function -- with a saddle at $P$, a local minimum at $Q$, and a local maximum at $R$, use $f\circ g$ instead of $f$, where $g$ is a diffeomorphism of the plane moving the given three distinct points $P,Q,R$ to the respective critical points $(0,\approx-8.54)$, $(0,\approx-1.14)$, $(0,\approx1.04)$ of $f$.

Such a diffeomorphism exists, because

• any three noncollinear points in $\R^2$ can be moved to any other three noncollinear points in $\R^2$ by a nonsingular affine transformation of $\R^2$ (which is of course a diffeomorphism);
• any three collinear points in $\R^2$ can be moved to some three noncollinear points in $\R^2$ by a diffeomorphism; say, three distinct collinear points $(0,y_1)$, $(0,y_2)$, $(0,y_3)$ in $\R^2$ can be moved to the three noncollinear points $(e^{y_1},y_1)$, $(e^{y_2},y_2)$, $(e^{y_3},y_3)$ by the diffeomorphism $h$ defined by the formula $h(u,v):=(u-e^v,v)$ for $(u,v)\in\R^2$;
• any diffeomorphism preserves critical points, local maxima points, local minima points, and saddle points.

Below are the calculations for $f$, in Mathematica:

$$f(x,y)=\frac{2}{x^2+(y-1)^2+1}-\frac{1}{x^2+(y+1)^2+1} \tag{1}\label{1}$$ 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 $(0,\approx1.04)$.

Below are the calculations, in Mathematica:

For any given three distinct points on the plane to be exactly the critical points (again, a saddle, a local minimum, and a local maximum) of a function, instead of $f$ use $f\circ g$, where $g$ is a diffeomorphism of the plane moving the given three distinct points to the critical points of $f$.

$$f(x,y)=\frac{2}{x^2+(y-1)^2+1}-\frac{1}{x^2+(y+1)^2+1} \tag{1}\label{1}$$ 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 $(0,\approx1.04)$.

For any given three distinct points on the plane to be exactly the critical points (again, a saddle, a local minimum, and a local maximum) of a function, use $f\circ g$ instead of $f$, where $g$ is a diffeomorphism of the plane moving the given three distinct points to the critical points of $f$.

Below are the calculations for $f$, in Mathematica:

$$f(x,y)=\frac{2}{x^2+(y-1)^2+1}-\frac{1}{x^2+(y+1)^2+1} \tag{1}\label{1}$$ 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 $(0,\approx1.04)$.

Below are the calculations, in Mathematica:

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 $\mathbb R^2\ni z\mapsto f(Az)$.

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.

$$f(x,y)=\frac{2}{x^2+(y-1)^2+1}-\frac{1}{x^2+(y+1)^2+1} \tag{1}\label{1}$$ 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 $(0,\approx1.04)$.

Below are the calculations, in Mathematica:

For any given three distinct points on the plane to be exactly the critical points (again, a saddle, a local minimum, and a local maximum) of a function, instead of $f$ use $f\circ g$, where $g$ is a diffeomorphism of the plane moving the given three distinct points to the critical points of $f$.