Partial derivatives of $g(x,y,z)=f(x-y,y-z,z-x)$. Let $f$ be two times differentiable function of variables $u,w,v$ and define $g(x,y,z)=f(x-y,y-z,z-x)$. Determine $\frac{\partial g}{\partial x}+\frac{\partial g}{\partial y}+\frac{\partial g}{\partial z}$ and express $\frac{\partial^2 g}{\partial x^2}$ in partial derivates of $f$.
This is from calc 3. I don't know how to solve this at all but I suspect there is some easy rule to do this. Help would be appreciated.

Chain rule, as proposed in the comments, is a sure path. However, it may generate quite bulky expressions, that is why it's judicious to prepare the ground by computing the differential of $g$ first. Renaming the variables as
$$
\begin{cases}
   t_1 = x-y \\
   t_2 = y-z \\
   t_3 = z-x
\end{cases}
$$
we can write
$$
\begin{array}{rcl}
\mathrm{d}g 
   &=&
   \mathrm{d}(f(x-y,y-z,z-x)) \\
   &=&
   f_1(t_1,t_2,t_3)\mathrm{d}t_1 + f_2(t_1,t_2,t_3)\mathrm{d}t_2 + f_3(t_1,t_2,t_3)\mathrm{d}t_3 \\
   &=&
   f_1(t_1,t_2,t_3)(\mathrm{d}x-\mathrm{d}y) \;+\\&& f_2(t_1,t_2,t_3)(\mathrm{d}y-\mathrm{d}x) \;+\\&& f_3(t_1,t_2,t_3)(\mathrm{d}z-\mathrm{d}x) \\
   &=&
   (f_1(t_1,t_2,t_3)-f_3(t_1,t_2,t_3))\mathrm{d}x \;+\\&& (f_2(t_1,t_2,t_3)-f_1(t_1,t_2,t_3))\mathrm{d}y \;+\\&& (f_3(t_1,t_2,t_3)-f_2(t_1,t_2,t_3))\mathrm{d}z
\end{array}
$$
where the indices denote partial derivatives. In consequence, we have :
$$
\frac{\partial g}{\partial x} + \frac{\partial g}{\partial y} + \frac{\partial g}{\partial z} = (f_1-f_3) + (f_2-f_1) + (f_3-f_2) = 0
$$
You can apply the same procedure to obtain $\frac{\partial^2g}{\partial x^2}$. Indeed, consdering the differental $g_x = \frac{\partial g}{\partial x}$, i.e.
$$
\begin{array}{rcl}
\mathrm{d}g_x 
   &=&
   \mathrm{d}(f_1(t_1,t_2,t_3)-f_3(z_1,z_2,z_3)) \\
   &=&
   f_{11}(t_1,t_2,t_3)\mathrm{d}t_1 + f_{12}(t_1,t_2,t_3)\mathrm{d}t_2 + f_{13}(t_1,t_2,t_3)\mathrm{d}t_3 \;-\\&& f_{31}(t_1,t_2,t_3)\mathrm{d}t_1 - f_{32}(t_1,t_2,t_3)\mathrm{d}t_2 - f_{33}(t_1,t_2,t_3)\mathrm{d}t_3 \\
   &=&
   \ldots \\
   &=&
   (f_{11}-2f_{13}+f_{33})\mathrm{d}x + (\ldots)\mathrm{d}y +(\ldots)\mathrm{d}z
\end{array}
$$
we find that
$$
\frac{\partial^2g}{\partial x^2} = f_{11}(t_1,t_2,t_3)-2f_{13}(t_1,t_2,t_3)+f_{33}(t_1,t_2,t_3)
$$
