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