Partial Derivatives of e^(xy) ln(1 + z^2)

Freeze two variables at a time

Nothing changes when you go from two variables to three: each partial freezes every variable except the one that moves. Here the function splits cleanly, with $x$ and $y$ living in the exponential and $z$ living in the logarithm, so each partial differentiates one factor and copies the other.

At $(1,1,1)$ the value is $e\ln 2 \approx 1.884$, the $x$ and $y$ slopes are both $e\ln 2 \approx 1.884$, and the $z$ slope is $\frac{2e}{2} = e \approx 2.718$.

The mistake: mixing factors across the split

• Writing $\frac{\partial f}{\partial z} = e^{xy}\ln\!\left(1+z^{2}\right)\cdot\frac{2z}{1+z^{2}}$ multiplies by the logarithm a second time. The logarithm is what you are differentiating, not a factor to keep.
• Writing $\frac{\partial f}{\partial x} = e^{xy}$ drops both the frozen factor $\ln(1+z^{2})$ and the inside derivative $y$.
• Writing a single expression that mixes all three partials. A function of three variables has three separate first partials, one per variable.

The plane $z = 0$ is the test. There $\ln(1) = 0$, so $f$ is identically $0$ on that whole plane and every slope must be $0$. The $x$ and $y$ partials carry the factor $\ln(1+z^{2})$ and vanish, and the $z$ partial carries the factor $2z$ and vanishes as well.

The gradient in three variables

Since $e^{xy} > 0$ and $\ln(1+z^{2}) \ge 0$, the function is never negative, and it equals $0$ exactly on the plane $z = 0$. The gradient is the zero vector on that entire plane, which is therefore a floor of minimum values rather than a set of isolated critical points.

Away from that plane the first two components share the factor $e^{xy}\ln(1+z^{2})$, so their ratio is just $y : x$, while the third component is controlled by how far $z$ is from $0$.