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.