How do I find the minimum value of this function? The question is: 

Consider the domain $\{ (x,y) \in \mathbb{R}^2 ; x \le y \}$
  The function $h$ is mapped from this domain to $R$ — the set of all real numbers: 
  $$h(x,y) = (x-2)^4 + (y-1)^4.$$ 
  Find the minimum value of this function. 

I found out the necessary partial derivatives — $h_x$ and $h_y$ and equated them to $0$ to get $x=2$ and $y=1$. Further, $h_{xx}$,$h_{yy}$ and $h_{xy}$ are all zero for these values of $x=2$ and $y=1$. 
So,H = $h_{xx}h_{yy} - h_{xy}^2$ is coming out to equal 0. How should I proceed now?
 A: Note that the critical point $(2,1)$ is NOT in the given domain $D=\{ (x,y) \in \mathbb{R}^2 ; x \le y \}$. It follows that the minimal value of $h$ is not in the interior of $D$. 
Now, in order to find the minimal value of $h$, you should consider the function $h$ restricted to the boundary of $D$, i.e. along the line $y=x$: 
$$f(x):=h(x,x)=(x-2)^4 + (x-1)^4$$
where $x\in \mathbb{R}$. We take the derivative of $f$ and find  its critical points:
$$f'(x)=4(x-2)^3 + 4(x-1)^3=4(2x-3)\underbrace{(x^2-3x+3)}_{>0}.$$
Therefore $x=3/2$ is the unique minimum point for $f$ and $(3/2,3/2)$ is the minimum point for $h$ with minimal value $h(3/2,3/2)=1/8$.
A: You have that  $h\geq0$ and that  $h=0$ is achieved at $(2,1) $. But that's outside your region!
Now partial derivatives will only detect local maxima/minima. When you determine all critical points, you can check for the largest and smallest values. 
To have all the critical points, you should always include the boundary of your region: in this case, the line  $y=x $. So, on this critical line, $g (x)=h (x,x)=(x-2)^4+(x-1)^4$.  So $g'(x)=4(x-2)^3+4 (x-1)^3$. The derivative is increasing and has a single zero at $x=3/2$, which gives the minimum.
A: I will be first fixing $x$ then find $y$ for which function is minimum. 
If $x\le1\Rightarrow y=1$ for minimum(minimum value of $(y-1)^4$ occur at $1$ and $y$ can take value $1$).But as we decrease $x$ , $(x-2)^4$ increase. So if $x\le1$ , then mininmum value of function occur at ($1,1$)=$1$
If $x\ge1$ then $x=y$ (because as we increase $y$ after $1$ function increase  , but $y\ge x$ ). For minimum vlaue of function in $x\ge1$ we will find minimum value of $(x-2)^4+(x-1)^4$=F(x) . 
F'($x$) = $4(x-2)^3+4(x-1)^3$ = $8(2x-3)(x^2-3x+3)$ , so F'($x$)$\ge0$ if $x\ge\frac{3}{2}$ and $\le0$ if $x\le\frac{3}{2}$ (as the quadratic have discriminate < 0 , hence always positive). Therefore minimum occur at $x=\frac{3}{2}.$
So , final answer is $\frac{1}{8}$. But i don't know how to solve by the method you were using.
