You should try to solve the equations. The six solutions wouldn't help. Following @Mann's idea, you can multiply the left hand sides of the three equations, and multiply the right hand sides of the three equations, then equate them:
$$xy\cdot yz\cdot xz =48 \lambda^3 xyz$$
then see if you can solve the $x,y,z$'s with the obtained information.
To make the solution complete, you should first observe that any one of $x,y,z=0$ is impossible, since one of them will imply the others and that contradict with your constraint.
So you get $xyz=48\lambda^3$. Notice your three original equations are in a pattern that is very consistent with this. For example, you can divide by $x$ to get $yz=\frac{48\lambda^3}{x}$ and plug this into the first equation. This gives you $x^2=24\lambda^2$. Doing the same thing for the other two equations, you can get similar equations for $y^2,z^2$. Now plug these into the constraint you will get an equation with respect to $\lambda^2$. Can you solve it from here?
We first find any critical point in the interior of the domain, which is $x^2 + y^2 < 1.$ So we have $f'(x, y, z) = (y + z, x - z, x - y)$ and $f'(x, y, z) = 0$ can only happen when $x = y = z$ and $y = -z,$ so $x = y = z = 0,$ but this is not in the interior of the domain. Hence, there are no critical points.
Consider now one part of the frontrier, the one defined by $x^2 + y^2 = z = 1.$ Hence we have to maximise $u(x, y) = f(x, y, 1) = x + xy - y$ with $x^2 + y^2 = 1.$ Set $g(x, y) = x^2 + y^2 - 1.$ Since $g'(x, y) = (2x, 2y),$ the derivative of $g$ is never zero and the Lagrange multipliers method apply. We have to solve $u'(x, y) = \dfrac{\lambda}{2} g'(x, y),$ which signifies $(1 + y, x - 1) = \lambda (x, y).$ Observe that $x = 1$ implies $y = 0$ and so we can consider the point $(1, 0)$ as one of the critical points in the restriction. Putting it aside, we have (by division), $\dfrac{1 + y}{x - 1} = \dfrac{x}{y}$ or else $y + y^2 = x^2 - x = 1 -y^2 - x$ or $x = 1 - 2y^2 - y.$ From the relation $x^2 + y^2 = 1$ we get $1 + 4y^4 + y^2 - 4y^2-2y+y^2 = 1,$ that is, $4y^4-2y^2-2y=0$ which leads to $y = 0$ (and hence $x = 1$) or else $2y^3-y-1=0.$ By inspection, we can factor $(y - 1)(2y^2+2y+1) = 0$ and the only solution is $y = 1$ with $x = -3,$ which is not in the restriction. Hence, we got only one point so far $(1, 0, 1).$
We finally tackle the other part of the restriction, which is $x^2 + y^2 - z = 0.$ Here we solve $f'(x, y, z) = \lambda (2x, 2y, -1)$ and this leads to (upon summing) $y + x = 2\lambda(x + y)$ and $y - x = \lambda.$ If $x = -y,$ then $z = 2x^2 \leq 1$ and we are optimising $h(x) = f(x, -x, 2x^2) = 4x^3 - x^2$ for $|x| \leq \dfrac{1}{2}$ and the optimisers for $h$ are $x = -\dfrac{1}{\sqrt{2}}$ and $x = \sqrt{2} - \dfrac{1}{2},$ this allow deducing the critical points for $f$ to be $\left(-\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}, 1\right)$ and $\left(\dfrac{1}{\sqrt{2}}, -\dfrac{1}{\sqrt{2}}, 1\right).$ If $x \neq -y,$ then $\lambda = \dfrac{1}{2}$ and $y = x + \dfrac{1}{2}.$ Again, substituting $x^2 + y^2 = z \leq 1$ leads to the point (details ommited) $\left(-\dfrac{1+\sqrt{7}}{4}, \dfrac{1 - \sqrt{7}}{4}, 1\right).$
To terminate the problem, recall the region is compact and evaluate in each of the four points.
Take my calculations with a grain of salt, I did them and checked them twice, but with calculus of several variables, I always commit arithmetic errors.
Best Answer
Your solutions are correct and Lagrange is simpler than polar coordinates.
Another simple method: from $g(x,y)=0$ we get $x^2=25-y^2$, hence $f(x,y)=75-3y^2-y^3$.
Now discuss the function $h(y)=75-3y^2-y^3$ on the intervall $[-5,5]$.