Partials of (x^2 - y^2)/(1 + x^2 y^2)

Apply the quotient rule, then collect terms

Hold $y$ fixed. The numerator $x^2 - y^2$ has $x$ derivative $2x$, and the denominator $1 + x^2y^2$ has $x$ derivative $2xy^2$.

Factor $2x$ out of the numerator and expand what is left: $1 + x^2y^2 - x^2y^2 + y^4$. The $x^2y^2$ terms cancel exactly.

The same computation in $y$ gives $-2y\left(1 + x^4\right)$ over the same squared denominator, with the minus sign inherited from the $-y^2$ in the numerator. At $(1,1)$ the two partials are $1$ and $-1$.

The mistake: stopping before the cancellation

The unsimplified quotient rule output is technically correct, but leaving it there hides the structure and makes the next step, a critical point hunt or a second derivative, far harder than it needs to be.

• Expand the numerator fully before you decide it cannot be simplified.
• Here $x^2y^2$ appears twice with opposite signs and disappears.
• The clean form shows immediately that $\frac{\partial f}{\partial x}$ has the same sign as $x$, since $1 + y^4 > 0$ always.

The other frequent error is a sign: $\frac{\partial}{\partial y}\left(x^2 - y^2\right) = -2y$, and forgetting that minus flips the whole answer.

What the signs tell you

Because $1 + y^4$ and $1 + x^4$ are always positive and the denominator is squared, the signs are decided entirely by $x$ and $-y$.

• $\frac{\partial f}{\partial x} > 0$ when $x > 0$ and $< 0$ when $x < 0$.
• $\frac{\partial f}{\partial y} < 0$ when $y > 0$ and $> 0$ when $y < 0$.
• Both vanish only at the origin, the single critical point.

That is exactly saddle behaviour: moving away from the origin along the $x$ axis increases $f$, and moving away along the $y$ axis decreases it. Restricted to the axes, $f$ behaves like $x^2$ in one direction and $-y^2$ in the other.

The denominator $1 + x^2y^2$ is at least $1$, so the function and both partials are defined and smooth on the entire plane, with no excluded curves to state.