Is stress dependent on strain? Stress-strain curves are drawn with strain on X-axis and stress on Y-axis. Usually the quantities placed on X-axis are independent quantities whereas the one placed on Y-axis are dependent. 
So which one is independent and why? 
 A: I think there is no correct answer to this question. There are cases where strain occurs without stress (i.e. heating of an unrestrained bar) and others where stress occurs without strain (i.e. heating of a fully restrained bar). 
If you are asking about uniaxial "engineering" stress-strain experimental curves, then the stress axis is obtained from reading the actuator force sensor and dividing by the initial area of the test bar. Therefore, we could say that this stress is independent of strain.
On the other hand, if you are looking at a "true" stress-strain curve; the actuator force is not divided by the initial area of the bar but by the current area of the bar. Therefore, the stress is a function of the strain.
A: So, note that here I'll use Einstein's convention on repeated indices with the usual tensor notation.
So, at the beginning, we all know that the forces $f_i$ are obtained from the stress tensor as follows
\begin{equation}
f_i=\frac{\partial\sigma_{ik}}{\partial x_k}
\end{equation}
Therefore, by the nature of the strain tensor $u_{ik}$ we have that the variation of work is
\begin{equation}
\delta W=-\sigma_{ik}\mathrm{d}u_{ik}
\end{equation}
Don't forget to do your calculations in order to reach this.
So, after some equation juggling you'll get to this result
\begin{equation}
\begin{aligned}
\mathrm{d}U&=T\mathrm{d}S+\sigma_{ik}\mathrm{d}u_{ik}\\
\mathrm{d}F&=-S\mathrm{d}T+\sigma_{ik}\mathrm{d}u_{ik}
\end{aligned}
\end{equation}
Therefore
\begin{equation}
\sigma_{ik}=\left(\frac{\partial F}{\partial u_{ik}}\right)_T=\left(\frac{\partial U}{\partial u_{ik}}\right)_S\tag{1}
\end{equation}
Then, jumping some passages (do the math) we can write the free energy as follows
\begin{equation}
\mathrm{d}F=Ku_{ll}\delta_{ik}\mathrm{d}u_{ik}+2\mu\left(u_{ik}-\frac{1}{3}\delta_{ik}u_{ll}\right)\mathrm{d}u_{ik}
\end{equation}
This should pop up if you have done the actual right math
Now you have enough to continue by yourself (note that it's only a simple derivation, and keep in mind that it's valid if and only if Hooke's law is valid in that case) [Use (1)].




source: Course on Theoretical Physics, Vol. 7: Theory of Elasticity, L. D. Landau && E. M. Lifshits, Chapter I, USSR Academy of Sciences.
A: It is possible to impose a known stress and the strain be the result. For example: the strain at the rope of a plumb line with a known weight.
Or the opposite, as when a guitar wire is strained by a known displacement, and the stress can be calculated from the pitch. 
But for 3D situations some displacements and some stresses, (or only some displacements) acting on a body are known. The remaining stresses and strains are calculated from the relation between stress and strain tensor. Normally they have to be estimated by finite elements method.
In a rolling mill for example, the imposed reduction of the thickness of a billet results in a stress tensor in the material, which is a function of the friction with the rolls. Depending on that tensor, the end product will elongate more (spreading less) or spreading more (elongating less).
