Partial Derivatives of ln(1 + x^2 + y^4)

Inner function down, its partial up

With $u = 1 + x^2 + y^4$, the outer derivative is $\frac{1}{u}$. The chain rule then multiplies by $\frac{\partial u}{\partial x} = 2x$ or $\frac{\partial u}{\partial y} = 4y^3$, so the inner function becomes the denominator and whichever inner partial you need becomes the numerator.

The $+1$ earns its place. It holds $u \ge 1$, so the logarithm is defined for every real pair $(x,y)$ and the denominator never vanishes. Drop it and the function falls to minus infinity at the origin.

The two numerators are not parallel

The $x$ term is squared and the $y$ term is a fourth power, so the numerators look nothing alike. Copying the shape of the first into the second, writing $2y$ where $4y^3$ belongs, is the error this function exists to expose. At $(1,2)$ the denominator is $18$, so the true $f_y$ is $\frac{32}{18} \approx 1.778$ while the copied version gives $\frac{4}{18} \approx 0.222$, out by a factor of $8$, which is $2y^2$.

The second slip is losing the numerator entirely and writing $\frac{1}{1 + x^2 + y^4}$. That is $\frac{\partial f}{\partial u}$, not $\frac{\partial f}{\partial x}$, and a sign test ends the argument: the true $x$ partial is negative for $x < 0$, while $\frac{1}{u}$ is positive everywhere.

The third is splitting the logarithm. $\ln(1 + x^2 + y^4)$ is not $\ln 1 + \ln x^2 + \ln y^4$. A logarithm turns a product into a sum, and there is no rule at all for a sum sitting inside it. The split version is also undefined whenever $x = 0$.

Much flatter along y than along x

Near the origin $f$ behaves like $x^2 + y^4$, and a fourth power is far flatter than a square. The partials put a number on it: at $(0.1,\,0)$ the $x$ partial is about $0.198$, while at $(0,\,0.1)$ the $y$ partial is about $0.004$, roughly fifty times smaller.

Both partials vanish only at the origin, since $2x = 0$ forces $x = 0$ and $4y^3 = 0$ forces $y = 0$. That single critical point is the global minimum, with $f = \ln 1 = 0$, because $u$ is smallest there and the logarithm is increasing.

Further out the picture reverses. Along the $y$ axis the partial $\frac{4y^3}{1 + y^4}$ grows to a peak and then decays like $\frac{4}{y}$, so the fourth power that made the surface flat near the origin is the same thing that makes it steep once $|y|$ passes $1$.