Applying Chain rule to $z = z(u, v) = f(x(u, v), y(u, v))$. If $z = z(u, v) = f(x(u, v), y(u, v))$ is a differentiable function, 
where $x = x(u, v) = u^2 − v^2$, $y = y(u, v) = 2uv$, show that
$$\frac{∂^2f}{∂x^2} +\frac{∂^2f}{∂y^2} =\frac{1}{4(u^2 + v^2)}\left(\frac{∂^2f}{∂u^2} + \frac{∂^2f}{∂v^2}\right)$$
 A: If we draw a dependency graph, the $u,v$ are independent at the bottom, $x=u^2-v^2$ & $y=2uv$ are in the middle, and $z=f(x,y)$ is at the top. Our neighboring/one-edge first derivatives are therefore:
$$\frac{\partial z}{\partial x}=f_x, \qquad \frac{\partial z}{\partial y}=f_y$$
$$x_u=\frac{\partial x}{\partial u}=2u, \qquad y_u=\frac{\partial y}{\partial u}=2v$$
$$x_v=\frac{\partial x}{\partial v}=-2v, \qquad y_v=\frac{\partial y}{\partial v}=2u$$
where the first line are merely shorthand notation for the first derivatives of $f$ (dispensing with the arguments $x,y$), and, by the chain rule, the two-edge first derivatives of $f$ are:
$$f_u=x_uf_x+y_uf_y=2uf_x+2vf_y$$
$$f_v=x_vf_x+y_vf_y=-2vf_x+2uf_y$$
Now remember that $z$ and $f$ are identified; the latter emphasizes its immediate dependency on $x$ & $y$. So the right hand side of what we want is actually not a "primitive" of $f$ and must be worked out. If we assume, or supply an argument why $f$ must in fact also have second derivatives, then the left hand side terms are indeed "primitives", in the sense that they would just be represented as $f_{xx}$ & $f_{yy}$, i.e. they are given along with $f$ (and its first derivatives and the mixed partials, $f_{xy}$ & $f_{yx}$ which play no explicit role as yet in this problem aside from assuming their equality). Thus, we can continue along these lines (exploiting the condensed subscript notation for partials):
$$\frac{\partial^2 f}{\partial^2 u}=f_{uu}
=2\left(uf_x+vf_y\right)_u
=2\left[f_x+u\left(f_x\right)_u+v\left(f_y\right)_u\right]$$
$$\frac{\partial^2 f}{\partial^2 v}=f_{vv}
=2\left(-vf_x+uf_y\right)_v
=2\left[-f_x-v\left(f_x\right)_v+u\left(f_y\right)_v\right]$$
Where so far we have used the product rule. Now we need the chain rule, as above, to calculate
$$(f_x)_u=x_uf_{xx}+y_uf_{xy}=2uf_{xx}+2vf_{xy}$$
$$(f_y)_u=x_uf_{xy}+y_uf_{yy}=2uf_{xy}+2vf_{yy}$$
$$(f_x)_v=x_vf_{xx}+y_vf_{xy}=-2vf_{xx}+2uf_{xy}$$
$$(f_y)_v=x_vf_{xy}+y_vf_{yy}=-2vf_{xy}+2uf_{yy}$$
so that
$$f_{uu}=2f_x+4\left(u^2f_{xx}+2uvf_{xy}+v^2f_{yy}\right)$$
and similarly
$$f_{vv}=-2f_x+4\left(v^2f_{xx}-2uvf_{xy}+u^2f_{yy}\right)$$
which together yield (in compact, subscript notation) the desired result:
$$f_{uu}+f_{vv}=4\left(u^2+v^2\right)\left(f_{xx}+f_{yy}\right)$$
A: First step let us compute the first derivatives:
$$\frac{\partial{f}}{\partial{u}}=\frac{\partial{f}}{\partial{x}}\frac{\partial{x}}{\partial{u}}+\frac{\partial{f}}{\partial{y}}\frac{\partial{y}}{\partial{u}}$$
$$\frac{\partial{f}}{\partial{v}}=\frac{\partial{f}}{\partial{x}}\frac{\partial{x}}{\partial{v}}+\frac{\partial{f}}{\partial{y}}\frac{\partial{y}}{\partial{v}}$$
And this gives
$$\frac{\partial{f}}{\partial{u}}=2u\frac{\partial{f}}{\partial{x}}+2v\frac{\partial{f}}{\partial{y}}$$
$$\frac{\partial{f}}{\partial{v}}=-2v\frac{\partial{f}}{\partial{x}}+2u\frac{\partial{f}}{\partial{y}}$$
And you derive again using the same rule...
