Gradient of x^3 y - x y^3: Full Worked Solution

Every term survives both partials, so work term by term

Neither term is free of $x$ or free of $y$, so nothing drops out. Handle one term at a time. For $\frac{\partial f}{\partial x}$, treat $y$ as a fixed number, which makes $y$ the coefficient of $x^3$ and $-y^3$ the coefficient of $x$.

For $\frac{\partial f}{\partial y}$, the roles reverse. Now $x^3$ multiplies $y$ and $-x$ multiplies $y^3$.

At $(2,1)$: the first component is $3(4)(1) - 1 = 11$ and the second is $8 - 3(2)(1) = 2$, so $\nabla f(2,1) = \langle 11, 2\rangle$.

The mistake: moving the exponent to the wrong variable

The classic wrong second component is $3x^2 - 3xy^2$, from differentiating $x^3 y$ with respect to $y$ and pulling the exponent off $x^3$ out of muscle memory. The exponent that moves is the one attached to the variable you are differentiating.

In $x^3 y$ the $y$ carries an invisible exponent of $1$. Bringing it down and reducing gives $x^3 \cdot 1 \cdot y^0 = x^3$, so the whole $x^3$ rides along untouched.

• $\frac{\partial}{\partial y}\left(x^3 y\right) = x^3$, not $3x^2$ and not $3x^2 y$.
• Watch the minus sign on $-xy^3$: it stays with the term, giving $-3xy^2$ in the $y$ partial and $-y^3$ in the $x$ partial.
• Symmetry check: swapping $x$ and $y$ turns $f$ into $-f$, and it also swaps the two gradient components with a sign flip. Both answers pass that test.

Where the gradient is zero

Factor each component. The first is $y\left(3x^2 - y^2\right)$ and the second is $x\left(x^2 - 3y^2\right)$, and both must vanish at once.

Suppose $y = 0$. Then the second equation becomes $x^3 = 0$, forcing $x = 0$. Suppose instead $3x^2 = y^2$. Substituting into the second component gives $x\left(x^2 - 9x^2\right) = -8x^3 = 0$, so again $x = 0$ and then $y = 0$. The origin is the only critical point.

The function itself factors as $f = xy(x - y)(x + y)$, so it is zero on four lines through the origin and alternates sign between them, eight sectors in all. That confirms the origin is a saddle rather than an extreme value.