Consider the function
$$ z(x,y) = y^3 + 2xy^2 - 3x^2y.$$
Determine the tangent line through $(x,y)=(2,1)$.
$$\begin{align}
\dfrac{z'_x(x,y)}{z'_y(x,y)}= \dfrac{2y^2-6xy}{3y^2+4xy-3x^2}
\end{align}$$
$$\begin{align}
\textrm{slope} &=- \dfrac{2\cdot 1^2-6\cdot 2\cdot 1}{3\cdot 1^2+4\cdot 2 \cdot 1-3\cdot 2^2}\\
&=-\dfrac{-10}{-1}\\
& = - 10.
\end{align}$$
The general form of a tangent line is $t(x)=ax+b$. Now it holds that $a=-10$.
In order to determine $b$ we use that the tangent line goes through $(x,y)=(2,1)$. Hence, $t(2)=-10\cdot 2 + b=1$, which results in $b=21$.
Hence, the tangent line is given by $t(x)=-10x+21$.
