 Solution

Inside the triangle, both $x$ and $y$ are nonzero. Thus $\nabla f=0$ reduces to $1-2x+y=0$ and $1-x+2y=0$, whose unique solution is $(1/3,-1/3)$. At that pointThe Hessian matrix has positive eigenvalues $1$ and $1/3$, so this is a strict local minimum. It is also the minimum on the closed triangle: $f=0$ on the boundary, $f<0$ inside, and there is no other interior stationary point. Therefore