Partial Derivatives of (x^2 + y^2 + 1)^(3/2)

Power rule outside, inner partial after

Treat the bracket as a single block $u = x^2 + y^2 + 1$. The power rule on $u^{3/2}$ gives $\frac{3}{2}u^{1/2}$, and then the chain rule multiplies by $\frac{\partial u}{\partial x} = 2x$.

The $\frac{3}{2}$ and the $2x$ multiply to $3x$, which is why the tidy form carries no fractions at all. The $y$ partial follows the same route.

Where the fractional exponent goes wrong

The first error is dropping the inner partial and writing $\frac{3}{2}\sqrt{x^2+y^2+1}$. That is $\frac{\partial f}{\partial u}$, not $\frac{\partial f}{\partial x}$, and it is wrong by the factor $2x$. At $(1,1)$ the bracket is $3$, so the true $f_x = 3\sqrt{3} \approx 5.196$ while the truncated version gives $\frac{3}{2}\sqrt{3} \approx 2.598$, exactly half.

The second error is distributing the exponent across the sum: $(x^2+y^2+1)^{3/2}$ is not $x^3 + y^3 + 1$. Exponents distribute over products, never over sums, and at $(1,1)$ the two differ by $5.196$ against $3$.

The third is the exponent arithmetic. The power rule subtracts $1$, and $\frac{3}{2} - 1 = \frac{1}{2}$, so what survives is a square root. Leaving the exponent at $\frac{3}{2}$, or flipping it to $-\frac{1}{2}$ as if the power were negative, both change how fast the answer grows.

A radial gradient and the role of the plus one

Both partials share the factor $3\sqrt{x^2+y^2+1}$, so $\nabla f = 3\sqrt{x^2+y^2+1}\,(x,\,y)$ points straight away from the origin. The surface is a bowl whose steepness grows without bound, and $(0,0)$ is the only critical point, with $f = 1$ there.

The $+1$ is doing real work. It keeps the base at least $1$, so the square root is never zero and the fractional power is smooth on the whole plane with no special case at the origin to argue about.