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.
I suggest to take geo method directly which is easy to understand. check above picture, you need to find the circle(center is fixed) max and min R in a triangle region that is very straightforward.
edit 1:(add more details)
$f(x,y)=(x-2)^2+y^2+5 \implies (x-2)^2+y^2=f(x,y)-5=R^2$
you want to find max and min of $f(x,y)$ is same to find max and min of $R$.
note the circle have to be inside the triangle so max of $R$ is $2$ ,min of R is $0$ which give $f_{max}=4+5=9, f_{min}=0+5=5$
if op insist on pure Lagrange multiplier method, it has much more things to write and you have three boundary equations . Anyway, if you don't like geo method, simply ignore it. it is true that we can't use this simple method if $f(x,y)$ is not the circle.
Best Answer
Maximum/minimum can occur in stationary points in the interior of the domain, in the boundary, or in points where $f$ is not differentiable. In this case, they will occur in the boundary or in the critical point that you mentioned. The function will attain a maximum on $(1/3,1/3)$ and a minimum all over the boundary($f=0$ on the boundary).