Find y' for e^y+xy=e, and evaluate y'(0)
Find $y'$ for $e^{y}+xy=e$, and evaluate $y'(0)$.
Explanation of Method
Locate the point first, then differentiate.
Step 1: Find $y$ at $x = 0$
Setting $x = 0$ collapses the product term: $e^{y} = e \;\Longrightarrow\; y = 1$.
So the point of interest is $(0,1)$.
Step 2: Differentiate both sides with respect to $x$
Chain rule on the exponential, product rule on $xy$: $e^{y}\,y' + y + x\,y' = 0$.
Step 3: Solve for $y'$
\[ y'\bigl(e^{y}+x\bigr) = -y \;\Longrightarrow\; y' = -\frac{y}{e^{y}+x}. \]
Step 4: Evaluate at $(0,1)$
\[ y'(0) = -\frac{1}{e^{1}+0} = -\frac{1}{e} \approx -0.36788. \]
Step 5: Verification
\[x = 0.001:\quad e^{y}+0.001y = e \;\Longrightarrow\; y \approx 0.999632\]
\[\frac{0.999632-1}{0.001} \approx -0.368 \approx -\frac1e \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = -\dfrac{y}{e^{y}+x}, \qquad y'(0) = -\dfrac{1}{e}\,}\]
Remark
Remark. Since $e^{y} \gt 0$, the denominator vanishes only where $x = -e^{y} \lt 0$; near the point of interest the implicit function theorem applies comfortably. Note that the relation cannot be solved for $y$ in elementary terms — it needs the Lambert $W$ function — yet the slope at a specific point costs three lines.