Answer
$$
xe^{y}=2-\ln2+\ln x; \quad [x=2 , \quad y=0 , \quad \frac{dx}{dt}=6]
$$
$$
\frac{d y}{d t}=-\frac{3}{2}
$$
Work Step by Step
$$
xe^{y}=2-\ln2+\ln x; \quad [x=2 , \quad y=0 , \quad \frac{dx}{dt}=6]
$$
Now, we can calculate $\frac{dy}{dt}$ by implicit differentiation, use the product rule as follows:
$$
\begin{aligned}
e^{y} \frac{d x}{d t}+x e^{y} \frac{d y}{d t} &=0+\frac{1}{x} \frac{d x}{d t} \\ x e^{y} \frac{d y}{d t} &=\left(\frac{1}{x}-e^{y}\right) \frac{d x}{d t} \\ \frac{d y}{d t} &=\frac{\left(\frac{1}{x}-e^{y}\right) \frac{d x}{d t}}{x e^{y}} \\
&=\frac{\left(1-x e^{y}\right) \frac{d x}{d t}}{x^{2} e^{y}} \\
& \quad \text { subistituting } \left[\begin{array}{c}{x=2 , \quad y=0 , \quad \frac{dx}{dt}=6 } \end{array}\right] , \\
&=\frac{\left[1-(2) e^{0}\right](6)}{2^{2} e^{0}} \\ &=\frac{-6}{4} \\
&=-\frac{3}{2} . \end{aligned}
$$
