We know that, if we have a surface $z=f(x,y)$ with Euclidean space being ambient manifold, the induced metric is as follows (in matrix form):
$$g=\begin{bmatrix}
1+\left ( \frac{\partial f(x,y)}{\partial x} \right )^2 & \frac{\partial f(x,y)}{\partial x}\frac{\partial f(x,y)}{\partial y} & \\\ 
\frac{\partial f(x,y)}{\partial x}\frac{\partial f(x,y)}{\partial y} & 1+\left ( \frac{\partial f(x,y)}{\partial y} \right )^2 & 
\end{bmatrix}$$


Now, consider a problem of prescribing an induced metric, i.e. assume we have a metric


$$h=\begin{bmatrix}
a(x,y) & b(x,y)\\\ 
b(x,y) & c(x,y)
\end{bmatrix}$$


The question is, when metric $h$ is an induced metric, i.e. when there exist a surface $z=f(x,y)$ for which an induced metric is $h$?


An example would be a metric
$$h=\begin{bmatrix}
2 & 1\\\ 
1 & 2
\end{bmatrix}$$


for which there exist a surface $z=-(x+y)$ with an induced metric being
$$g=\begin{bmatrix}
2 & 1\\\ 
1 & 2
\end{bmatrix}$$


An obvious obstruction is that the following identity has to be satisfied by the givien metric:
$$a(x,y)c(x,y)-a(x,y)-c(x,y)+1=b^2(x,y).$$
This identity follows from requirement that 
$$\frac{\partial f(x,y)}{\partial x}\frac{\partial f(x,y)}{\partial y}=b(x,y).$$


Ths problem occured when toying with several examples of positive definite Hessians (with minus sign), i.e. when (minus) Hessian for some other surface $z=\phi(x,y)$ is positive definite. For example, if $z=-(x^2+xy+y^2)$, then the minus Hessian is
$$h=\begin{bmatrix}
2 & 1\\\ 
1 & 2
\end{bmatrix}$$
This matrix can be taken as a first fundamental form of the surface $z=-(x+y)$, which has a zero Gaussian curvature. It seems that often when negative Hessian is taken as a first fundamental form, Gaussian curvature (for a surface with induced metric being equal to this negative Hessian) is equal to zero. I'm not claiming that this is a general phenomena, it's just an interesting observation (it is important from the point of view of optimization as Hessian often play a significant role there).


I looked for litereture on "prescribed metric" problem, but found none. Probably I wasn't looking in the right place. **A link to a paper, of a monograph considering this problem would be appreciated**.


I'm familiar with the problem of prescribed curvature. Probably one possible way would be to look for prescribed Ricci curvature tensor problem. But at this stage I think it would be an overkill. And maybe the class of possible solutions (taking into consideration the obstructions) is way too small and simple to be of value: it looks like negative Hessians for quadratic surfaces might a class of solutions.



Prescribing an induced metric
Assume $b\ne0$.


Then you know $\alpha=\tfrac{\partial f}{\partial x}$ and $\beta=\tfrac{\partial f}{\partial y}$ up to sign. 
You should check $$\frac{\partial \alpha}{\partial y}=\frac{\partial \beta}{\partial x}.$$ 
If this is true you can restore $f$ by integrating.


