Does $\left(\frac{\partial f}{\partial x}\right)^2=\frac{\partial^2f}{\partial x^2}$ This may be an obvious question but I'm just not thinking straight, thanks
The answer must be no
 A: Of course not !
Take for exemple 
$$f(x,y)=x,$$
then
$$\left(\frac{\partial f}{\partial x}\right)^2=1,$$
but
$$\frac{\partial^2 f}{\partial x^2}=0.$$
A: As others have noted, nope. But let's explore for which functions it is true. If the function is not linear, it has a second derivative that is not always zero, so the first step here is valid. (And if the function is linear, we can work out that it would have to be a constant function.)
$$
\begin{align}
\frac{\partial^2f}{\partial x^2}
&=\left(\frac{\partial f}{\partial x}\right)^2\\
\left(\frac{\partial f}{\partial x}\right)^{-2}\frac{\partial^2f}{\partial x^2}
&=1&\text{anitdifferntiate on both sides}\\
-\left(\frac{\partial f}{\partial x}\right)^{-1}
&=x+C_1\\
\frac{\partial f}{\partial x}
&=\frac{-1}{x+C_1}&\text{anitdifferntiate on both sides}\\
f(x)&=-\ln\left|x+C_1\right|+C_2(x)
\end{align}
$$
where $C_2$ is some function that is a constant value on $\left(-C_1,\infty\right)$ and a possibly different constant value on $\left(-\infty,-C_1\right)$. You could also write it as $$f(x)=-\ln\left|x+C\right|+D\frac{x+C}{|x+C|}+E$$
A: Absolutely not. Pick any function you like and work out the RHS and LHS, and you will almsot always get two different solutions.
A: This is not true. To see this, let's consider an example (there are simpler examples...)
$$
f(x,y) = x^2y^2
$$
Then
$$
\frac{\partial f}{\partial x} = 2xy^2 \\
\frac{\partial^2 f}{\partial x^2} = 2y^2.
$$
And
$$\left(\frac{\partial f}{\partial x}\right)^2 = (2xy^2)^2 = 4x^2y^4. 
$$
A: No. Let $f(x,y)=x$. Verification is left to you.
