= Solution
The <gradient> vanishes when $4x+3y-1=0$ and $3x+2y-1=0$. Subtracting these equations gives $x+y=0$, so the unique <critical point> is $\boxed{(x,y)=(1,-1)}$, with $f(1,-1)=0$.
The <Hessian matrix> is
$$
H=\begin{pmatrix}4&3\\3&2\end{pmatrix},\qquad \det H=8-9=-1<0.
$$
It is indefinite, so the <critical point> is a \b[<saddle point>]. Explicitly, around that point the change is exactly $2u^2+3uv+v^2$: it is positive for $v=0,u\ne0$ and negative for $v=-3u/2,u\ne0$. Thus arbitrarily close points have both larger and smaller values.
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