Linearization of an implicitly defined function. Problem:
Given the equation: $xz^{2}+y^{2}z^{5}=19$
Also given: (3,4,1) is a solution to the equation.  This point is not the only solution. 
1) Find dz/dx and dz/dy (through implicit differentiation) evaluated at (3,4).
2) Find the linearization L(x,y) of z(x,y) at (3,4)
3)  Use L to approximate z(3.01, 4.02).  Plug this (approximate) result into the given equation to see if it satisfies it.

My attempted solution:
1) dz/dx = $(-z^{2})/(2xz+5y^{2}z^{4})$
dz/dx evaluated at (3,4) = $(-z^{2})/(6z+80z^{4})$
dz/dy = $(-2yz^{5})/((2xz+5y^{2}z^{4})$
dz/dy evaluated at (3,4) = $(-8z^{5})/(6z+80z^{4})$
2) Formula for linearization: L(x) = f(a) + Df(a) (x-a)
f(a) = $3z^{2} + 16z^{5}=19$
$(z-1)(16z^{4} + 16z^{3} + 16z^{2} +19{z} +19)=0$
z=1 is the only real solution
I'm not entirely sure I am solving f(a) correctly.  Up until now, I have only dealt with explicitly defined functions.
Df(a) = Dz(a)[dz/dx evaluated at a  dz/dy evaluated at a]
Dz= \begin{bmatrix}
(-z^{2})/(6z)+80z^{4}),(-8z^{5})/((6z)+80z^{4})&\end{bmatrix}
(x-a) = 
\begin{bmatrix}
x-3,y-4\\ \end{bmatrix} (Transposed so that it is a column vector).
L(x) = $1 + (x-3)*((-z^{2})/(6z+80z^{4})+(y-4)*((-8z^{5})/(6z+80z^{4}))$
3) $1+((-0.01z^{2})/(6z+80z^{4})) + ((-0.16z^{5}/(6z+80z^{4}))$
Is this the approximation for z(3,4)?  I'm not sure how to proceed from here.  What do I plug into the original equation to see that it approximately satisfies the equation?
Thanks!  Sorry about the formatting. I tried to tidy up the formatting a bit.  I'm not too familiar with latex though.
 A: 1) Your differentiation is correct. Plug $z=1$ to have numerical values.
2) You have a function in the form of
$$L(x,y,z)=xz^2+y^2z^5-19=0$$
The linearization around $(x_0,y_0,z_0)$ is as follows
$$\hat L(x,y,z)=L(x_0,y_0,z_0)+\frac{\partial L}{\partial x}|_{x_0,y_0,z_0}(x-x_0)+\frac{\partial L}{\partial y}|_{x_0,y_0,z_0}(y-y_0)+\frac{\partial L}{\partial z}|_{x_0,y_0,z_0}(z-z_0)=0$$
By some manipulation and since $L(x,y,z)=0$ by definition
$$z=z_0-\frac{\frac{\partial L}{\partial x}|_{x_0,y_0,z_0}(x-x_0)+\frac{\partial L}{\partial y}|_{x_0,y_0,z_0}(y-y_0)}{\frac{\partial L}{\partial z}|_{x_0,y_0,z_0}}$$
Here you can find partial derivatives and plug $(x_0,y_0,z_0)=(3,4,1)$ to have a linear eqaution.
$$\frac{\partial L(3,4,1)}{\partial x}=z^2=1$$
$$\frac{\partial L(3,4,1)}{\partial y}=2yz^5=8$$
$$\frac{\partial L(3,4,1)}{\partial z}=2xz+5y^2z^4=86$$
$$\Rightarrow z=1-\frac 1{86}(x-3)-\frac 8{86}(y-4)$$
3) For $(x,y)=(3.01,4.02)$ the linear equation gives for $z$
$$z=1-\frac 1{86}(0.01)-\frac 8{86}(0.02)=0.998$$
If you plug it into original equation
$$L(3.01,4.02,0.998)=-0.00259$$
which is the linearization error.
