Compute partial derivatives: 

$f_x=2x-2,\; f_y=4y+4$.

Set $f_x=f_y=0 \Rightarrow 2x-2=0,\; 4y+4=0 \Rightarrow (x,y)=(1,-1)$.

Hessian: $H=\begin{pmatrix}2 & 0\\ 0 & 4\end{pmatrix}$ (positive definite since $2>0$ and $\det=8>0$).

Therefore, $(1,-1)$ is a strict (global) minimum. Also, by completing squares: $f(x,y)=(x-1)^2+2(y+1)^2-5 \Rightarrow f_{\min}=-5$ at $(1,-1)$.

Answer: Critical point $\boxed{(1,-1)}$; nature: $\boxed{\text{minimum}}$; minimum value $\boxed{-5}$.