Not sure how to interpret this question or where to start.
$\text{Assuming that the equation}$ $$F(x,y,z) = 0$$ $\text{defines} z \text{implicitly as a differentiable function of} \: x \: \text{and} \: y \: \text{and that}$ $$F_{zx} = F_{xz}$$ $\text{show that}$ $$\frac{\partial ^2 z}{\partial x^2} = \frac{-(F_{z})^2 F_{xx} + 2F_{z}F_{x}F_{xz} - (F_{x})^2 F_{zz}}{(F_{z})^3}.$$
I have no idea how to use the given equation to imply that. I know how to implicity differentiate functions but when they actually give a function...
Also, from their definition, it means that this is true right?
$z \equiv z(x,y)$
(We haven't learn the implicit function theorem btw).
A heuristic approach: from $F(x,y,z)=0$, you differentiate in terms of $x$ and $y$ repsectively (just treat $z=z(x,y)$) to get \begin{align} F_x+F_zz_x &= 0 \\ F_y+F_zz_y &=0 \end{align} and you solve $(z_x,z_y)=-\frac1F_z(F_x,F_y)$.
Then you differentiate $F_x+F_zz_x=0$ wrt $x$ again to obtain $$F_{xx} + F_{zx}z_x + z_x(F_{xz}+F_{zz}z_x)+F_zz_{xx}=0$$ and you get $z_{xx}$. It turns out $z_y$ is unused, but in general even if you only need the second partial derivative in a particular defining variable, you will still have to differentiate in all the defining variables as all first order partial derivatives can arise later on.