Answer
$$
y\ln x+xe^{y}=1, \quad [x=1 , \quad y=0 , \quad \frac{dx}{dt}=5]
$$
$$
\frac{d y}{d t}=-5
$$
Work Step by Step
$$
y\ln x+xe^{y}=1, \quad [x=1 , \quad y=0 , \quad \frac{dx}{dt}=5]
$$
Now, we can calculate $\frac{dy}{dt}$ by implicit differentiation, use the product rule as follows:
$$
\begin{aligned} \frac{d}{d t}(y \ln x)+\frac{d}{d x}\left(x e^{y}\right) &=\frac{d}{d t}(1) \\
\ln x \frac{d y}{d t}+\frac{y}{x} \frac{d x}{d t}+e^{y} \frac{d x}{d t}+x \cdot e^{y} \frac{d y}{d t} &=0 \\
\left(\ln x+x e^{y}\right) \frac{d y}{d t}+\left(\frac{y}{x}+e^{y}\right) \frac{d x}{d t} &=0 \\
\left(\ln x+x e^{y}\right) \frac{d y}{d t} &=-\left(\frac{y}{x}+e^{y}\right) \frac{d x}{d t} \\
\frac{d y}{d t} &=-\frac{\left(\frac{y}{x}+e^{y}\right) \frac{d x}{d t}}{\ln x+x e^{y}} \\
&=-\frac{\left(y+x e^{y}\right) \frac{d x}{d t}}{x \ln x+x^{2} e^{y}} \\
& \quad \text { subistituting } \left[\begin{array}{c}{x=1, \\ y=0 ,\\ \frac{dx}{dt}=5 } \end{array}\right]\\
\frac{d y}{d t}&=-\frac{\left[0+(1) e^{0}\right](5)}{(1) \ln 1+1^{2} e^{0}} \\
&=-5
\end{aligned}
$$
