Partials of sin(x) / (2 + cos(y))

Only one of the two partials needs the quotient rule

With $y$ held fixed the denominator is a constant, so the function is a constant multiple of $\sin x$ and no quotient rule is needed.

With $x$ held fixed the numerator is the constant. Write $f = \sin x\,(2+\cos y)^{-1}$ and use the power rule, which is faster than the full quotient rule here.

Two minus signs multiply to a plus, one from the power rule and one from the derivative of cosine. At the origin the partials are $\frac{1}{3}$ and $0$.

The mistake: one minus sign too many or too few

Almost every wrong answer for $\frac{\partial f}{\partial y}$ here is off by a sign. Track the two sources separately. Rewriting as a negative power contributes a factor $-1$. Differentiating the inside $2 + \cos y$ contributes $-\sin y$. Their product is positive.

• Verify numerically at $x = \frac{\pi}{2}$, $y = \frac{\pi}{2}$: the denominator is $2$, so $f = \frac{1}{2}$. Nudging $y$ up to $1.6$ makes $\cos y$ smaller, the denominator smaller, and $f$ larger, so $\frac{\partial f}{\partial y}$ must be positive there. The formula gives $\frac{1\cdot 1}{4} = 0.25 > 0$.
• A second slip is applying the quotient rule to $\frac{\partial f}{\partial x}$ and writing a numerator term for the derivative of the denominator. That term is zero, since $2 + \cos y$ has no $x$ in it.

Why the denominator is safe

Since $\cos y$ lies between $-1$ and $1$, the denominator $2 + \cos y$ lies between $1$ and $3$. It never touches zero, so the function and both partials are smooth on the entire plane. That is the reason for the $2$: with $\cos y$ alone in the denominator the function would blow up along infinitely many lines.

Critical points come from setting both partials to zero. The first needs $\cos x = 0$, so $x$ is an odd multiple of $\frac{\pi}{2}$, and there $\sin x = \pm 1$. The second then needs $\sin y = 0$, so $y$ is a multiple of $\pi$. Those points are the peaks and valleys of the surface, with heights $\pm 1$ and $\pm\frac{1}{3}$.