Second partial derivatives of sin(xy)

Chain rule first, then decide whether you also need the product rule

The inside is $u = xy$. Differentiating in $x$ with $y$ held fixed, the inner derivative is $\frac{\partial u}{\partial x} = y$, so $f_x = y\cos(xy)$. By symmetry $f_y = x\cos(xy)$.

For $f_{xx}$, differentiate $y\cos(xy)$ in $x$ again. The factor $y$ out front is a constant here, so only the chain rule is in play and it contributes a second factor of $y$.

The mixed partial is the interesting one. Differentiating $y\cos(xy)$ in $y$ hits $y$ in two separate places, the leading factor and the inside of the cosine, so the product rule is required.

The mistake: dropping the cos(xy) term

By far the most common wrong answer is $f_{xy} = -xy\sin(xy)$. It comes from differentiating the inside of the cosine and forgetting that the leading $y$ also depends on $y$. The result fails an easy test: at the origin the correct mixed partial is $\cos 0 = 1$, while the wrong version gives $0$.

A quick way to stay honest is to ask, before differentiating, how many times the variable appears in the expression. In $y\cos(xy)$ the variable $y$ appears twice, so the answer must have two terms. In $y\cos(xy)$ the variable $x$ appears once, so $f_{xx}$ has a single term and no product rule.

Check Clairaut symmetry by taking the other route. Starting from $f_y = x\cos(xy)$ and differentiating in $x$ gives $\cos(xy) + x(-y\sin(xy)) = \cos(xy) - xy\sin(xy)$, the same expression.

What the Hessian says at the origin

Both first partials vanish at $(0,0)$, so the origin is a critical point. Substituting there gives $f_{xx} = 0$, $f_{yy} = 0$ and $f_{xy} = 1$.

A negative discriminant means a saddle. That matches the picture: along the line $y = x$ the function is $\sin(x^2)$, which rises away from the origin, while along $y = -x$ it is $-\sin(x^2)$, which falls. The pure second partials being zero is what forces the classification onto the mixed partial.