There are two ways to proceed. One is to use implicit differentiation as before, but using partial derivatives. Whenever we differentiate a function of \(z\text{,}\) we multiply by the appropriate partial derivative of \(z\text{.}\) The other option is to use the formula derived above. We will use the first method for the \(x\) derivative, and the second for \(y\text{.}\)
We first take the partial derivative of both sides of
Equation (13.5.3) with respect to
\(x\text{:}\)
\begin{align*}
\frac{\partial}{\partial x}(x^2yz^3-\sin(x-3z)+4xy^2-3yz) \amp = 0\\
2xyz^3 + x^2y(3z^2)\plz{z}{x}-\cos(x-3z)\left(1-3\plz{z}{x}\right)+4y^2-3y\plz{z}{x} \amp = 0\text{.}
\end{align*}
Note that we treated \(y\) as a constant, since the derivative is with respect to \(x\text{.}\) Next, we collect terms:
\begin{equation*}
\plz{z}{x}\left(3x^2yz^2+3\cos(x-3z)-3y\right) = -2xyz^3+\cos(x-3z)-4y^2\text{.}
\end{equation*}
Lastly, we solve for \(\plz{z}{x}\text{:}\)
\begin{equation*}
\plz{z}{x} = \frac{-2xyz^3+\cos(x-3z)-4y^2}{3x^2yz^2+3\cos(x-3z)-3y}\text{.}
\end{equation*}
For the \(y\) derivative, we will use the result given above. Setting \(f(x,y,z) = x^2yz^3-\sin(x-3z)+4xy^2-3yz\text{,}\) we have \(\plz{z}{y} = -\frac{f_y(x,y,z)}{f_z(x,y,z)}\text{.}\) Therefore,
\begin{equation*}
\plz{z}{y} = -\frac{x^2z^3+8xy-3z}{3x^2yz^2+3\cos(x-3z)-3y}\text{.}
\end{equation*}
The second method certainly seems simpler! The reader is invited to try each part with the other method, and compare answers.
Finally, we consider the problem of the tangent plane. First, we check that the point \((3,0,1)\) is indeed on the surface: \(f(3,0,1)=0\text{,}\) as required. Next we note that \(z=1\) is given to us from this point. So if \(f(x,y,z)=c\) implicitly defines the graph \(z=g(x,y)\text{,}\) then we must have \(g(3,0)=1\text{.}\) Next, we have
\begin{align*}
g_x(3,0) \amp = \plzoa{z}{x}{(3,0)} = \frac{0+1-0}{0+3-0}=\frac13 \\
g_y(3,0) \amp = \plzoa{z}{y}{(3,0)} = -\frac{9+0-3}{0+3-0} = -2\text{.}
\end{align*}
The equation of the tangent plane is therefore
\begin{equation*}
z = g(3,0)+g_x(3,0)(x-3)+g_y(3,0)(y-0) = 1+\frac13(x-3)-2y\text{.}
\end{equation*}