I was surprised to find this question abandoned, so I thought I'd finish it.
It is clear that Farmor has already noted that setting $ \ \frac{\partial f}{\partial x} = y \ $ and $ \ \frac{\partial f}{\partial y} = x \ $ equal to zero locates the critical point at $ \ (0,0) \ , $ at which $ \ f(0,0) = 0 \ . $ The function is also zero at the boundaries of the triangle and at all three vertices (thus responding to Daryl's remark, although OP may well have already done that if that is what was meant by having inspected the boundaries). So the minimum value of the function in the region is zero, but that is not localized to a "critical point".
As for the boundary on $ \ x + y = 1 \ $ , the single-variable approach is to write $ \ f(x,y) = xy \ $ as $ \ f(x) = x \cdot (1-x) = x - x^2 \ , $ which has its maximum at $ \ f(\frac{1}{2}) = \frac{1}{4} \ $ (so identified since $ \ f'(\frac{1}{2}) = -2 \cdot \frac{1}{2} < 0 \ $ ) ; this can be found even without calculus by noting the properties of the "downward-facing" parabola $ \ y = x - x^2 \ . $
Alternatively, we can use Lagrange multipliers with the function $ \ f(x,y) = xy \ $ under the constraint $ \ g(x,y) = x + y - 1 \ , $ producing the equations
$$\begin{array}{cc}f_x \ = \ \lambda \ \cdot \ g_x\\f_y \ = \ \lambda \ \cdot \ g_y\end{array} \ \Rightarrow \ \begin{array}{cc}x \ = \ \lambda \ \cdot \ 1\\y \ = \ \lambda \ \cdot \ 1\end{array} $$
$$ \Rightarrow \ \lambda \ = \ x \ = \ y \ = \ \frac{1}{2} \ . $$
So we find what proves to be a local and absolute maximum, $ \ f(\frac{1}{2} , \frac{1}{2}) = \frac{1}{4} . $ The standard discriminant $ \ f_{xx} \cdot f_{yy} - ( f_{xy} )^2 = 0 \cdot 0 - 1^2 \ $ is of no help, however, in characterizing critical points for this function.
In many extremal problems the set $S\subset{\mathbb R}^n$ on which the extrema of some function $f$ are sought is stratified, i.e., consists of points of different nature: interior points, surface points, edges, vertices. If an extremum is assumed in an interior point it comes to the fore as solution of the equation $\nabla f(x)=0$. An extremum which is at a (relative) interior point of a surface or an edge comes to the fore by Lagrange's method or via a parametrization of this surface or edge. Here (relative) interior refers to the following: Lagrange's method deals only with constrained points from which you can march in all tangent directions of the submanifold (surface, edge, $\ldots$) defined by the constraint(s), all the while remaining in $S$. Now at a vertex there are forbidden marching directions on all surfaces meeting at that vertex. If the extremum is taken on such a vertex it only comes to the fore if you have deliberately taken all vertices into your candidate list.
Now your $S_1$ is an arc in the plane with two endpoints. (The latter are not immediately visible in your presentation of $S_1$, but you have found them.) Your candidate list then should contain all relative interior points of the arc delivered by Lagrange's method plus the two boundary points.
The circle $S_2\!: \ x^2+y^2=1$ however has only "interior" points. The candidate list then contains only the points found by Lagrange's method.
Best Answer
The domain is not a triangle; it's an ellipse with semi-axes $\sqrt2$ and $1$. We have $\nabla(f)=(2x,2y)$, which is a critical point (zero) only at $(0,0)$, which is a minimum.