Show $(\partial^2 z / \partial x \partial y)^2 = \frac{\partial^2z}{\partial x^2} \cdot \frac{\partial^2z}{\partial y^2}$ for $z=ax + yf(a)+\phi(a)$.

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Show $(\frac{\partial^2 z}{ \partial x \partial y})^2 = \frac{\partial^2z}{\partial x^2} \cdot \frac{\partial^2z}{\partial y^2}$ for the implicitly defined $z(x,y)$ as $$ z=ax + yf(a)+\phi(a) $$ and $$ 0 = x + yf'(a) + \phi'(a). $$

How do we know what functions are dependent on others? For example, we can arrange the first equation to solve for $f(a)$. Does this imply $f(a)$ is in fact an implicitly defined function of $z,y,x,\phi$ as well as $a$?

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The equation

$$x + yf'(a) + \phi'(a) = 0$$

defines (implicitly) $a$ as a function of $x,y$ so $a = a(x,y)$. I'm assuming that $f$ and $\phi$ are two given differentiable functions.

Start by calculating

$$z_x \equiv \frac{\partial z}{\partial x} = a + a_x[x + y f'(a) + \phi'(a)]$$ $$z_y \equiv \frac{\partial z}{\partial y} = f(a) + a_y[x + y f'(a) + \phi'(a)]$$

Use the first equation above to simplify and then proceed to calculate $z_{xx}, z_{yy}$, $z_{xy}$ and $z_{yx}$. Combinding these equation will give you

$$z_{xy}z_{yx} = z_{xx} z_{yy}$$

and if $f,\phi$ are 'nice enough' functions then $z_{xy} = z_{yx}$ and the result follows.