Partial Derivatives of x^2 e^(-y)

Differentiate one factor, copy the other

With $y$ held fixed, $e^{-y}$ is a fixed positive number and the function is a constant times $x^{2}$, so the power rule finishes the job.

With $x$ held fixed, $x^{2}$ is the constant and you differentiate $e^{-y}$. The exponent $-y$ has derivative $-1$, and that is where the minus sign enters.

At $(2,0)$ the slopes are $2\cdot 2\cdot 1 = 4$ in $x$ and $-4$ in $y$. Same size, opposite signs, because at that point the growth in $x$ and the decay in $y$ happen to balance.

The mistake: losing the minus sign on e^(-y)

• Writing $\frac{\partial f}{\partial y} = x^{2}e^{-y}$ treats $e^{-y}$ like $e^{y}$. The chain rule on the exponent $-y$ contributes a factor of $-1$, so the true slope is negative wherever $x \neq 0$.
• Writing $\frac{\partial f}{\partial y} = -x^{2}e^{-y-1}$ or $-x^{2}ye^{-y}$ invents a power rule for the exponential. Only the exponent's derivative comes out front; the exponential itself is unchanged.
• Writing $2x\,e^{-y} - x^{2}e^{-y}$ glues the two partials into one expression. Each partial is a separate answer.

The sign check is easy to remember. For a fixed $x \neq 0$, raising $y$ shrinks the factor $e^{-y}$ and lowers the surface, so $\frac{\partial f}{\partial y}$ must be negative. The correct formula, $-x^{2}e^{-y}$, is negative there for every $y$.

A whole line of flat points

Both partials carry a factor of $x$, so both vanish exactly when $x = 0$. This surface does not have an isolated critical point; it has the entire $y$-axis, where $f = 0$.

That matches the shape. Since $x^{2} \ge 0$ and $e^{-y} > 0$, the surface never dips below the plane $z = 0$ and touches it along that whole line, a valley floor rather than a single dip. The second derivative test is inconclusive along such a line, and you read the behaviour off the formula instead.