Find double derivative using implicit derivation - mulitvariable calculus Short question:
$z(x,y)$ is a function that is implicitly defined by the equation
$$4x+3y+5z+4\cos(4z)+3=0$$
in the neigbourhood of the point $p=(-5\pi/16, 1/3, \pi/4)$
I am going to calculate $\partial$ in $p$.

$F$ differentiated with respect to $x$ is $4$ and with respect to $y$ is $3$. 
Differentiated with respect to $z$ is $5-16\sin(4z)$
Then we get 
$$\frac{\partial z}{\partial x}=-4/5$$
and
$$\frac{\partial z}{\partial y}=-3/5 $$
by plugging in $z=\pi/4$
But now I don't know how to calculate $\frac{\partial^2z}{\partial x \partial y}$.
Can somebody help me?
 A: In computing these partials, we will assume $F_z \not = 0$. Thus, since $z = g(x,y)$, for a small neighborhood of the points $(x,y,g(x,y))$ we can choose a curve $c(t) \in \mathbb{R}^2$ such that:
\begin{align*} (x,y,g(x,y)) \in F( c(t)) \Rightarrow 0 &= \frac{d}{dt}(F \circ c)(t)\\&= \nabla F \cdot \frac{\partial}{\partial x} (x,y,g(x,y) \\&= (F_x, F_y, F_z)  \cdot (1,0, g_x) \\& \Rightarrow F_x + F_z g_x 
=0 \\& \frac{\partial z}{\partial x} = - \frac{F_x}{F_z} \end{align*} 
Here $F(x,y,z) = 4x+3y+5z+4\cos(4z)+3$. I'll leave you to the rest. Just know you should get zero.
A: For the higher derivatives you first note that $\frac{\partial z}{\partial x}(x,y) = -\frac{4}{5-16\sin(4z(x,y))}$. Now let's differentiate this expression with respect to $y$. After some work we should get that:
\begin{align*}
 \frac{\partial^2z}{\partial x  \partial y}(x,y) = -\frac{256 \frac{\partial z}{\partial y}(x,y) \cos(4z(x,y))}{(5-16\sin(4z(x,y)))^2}.
\end{align*}
Now remember from the first derivatives you already calculated that $\frac{\partial z}{\partial y}(x,y) = -\frac{3}{5-16\sin(4z(x,y))}$. You can now plug that into the expression for $\frac{\partial^2z}{\partial x  \partial y}(x,y)$.
Hope that helped.
