The Bernoulli IVP $$ x,y >0: \qquad y'(x)=\frac{y(x)}{x}+\sqrt{x^3y(x)^3}, \quad y(1)=1 $$ has the solutions $$ y_1(x) = \frac{36x}{(x^3 - 7)^2}, \qquad y_2(x) = \frac{36x}{(x^3 + 5)^2} \quad (x > 0) $$ which I found using the default substitution $z(x) := 1/\sqrt{y(x)}$. I am not sure that those are all solutions to this IVP.
Question: How many solutions does the Bernoulli IVP have in general?
Only one of these functions is the right solution. The answer comes in the solving process $$ z' + \frac{1}{2x}z = - \frac{1}{2}x^{3/2}$$
$$ (\sqrt{x}z)' = -\frac{x^2}{2}$$
$$ \sqrt{x}z = -\frac{x^3+c}{6} $$ $$ z = -\frac{x^3+c}{6\sqrt{x}} = \frac{1}{\sqrt{y}}$$
From this step, note that the RHS is positive, therefore $x^3 + c < 0$.
With the above condition in mind, and the fact that the solution is only defined for $x > 0$, we pick $c = -7$ and the domain is further restricted to $(0,\sqrt[3]{7})$
Sidenote: This is one of the cases where squaring gets you in trouble, such as $$ x = 2 $$ $$ x^2 = 4 $$ The second equation has 2 solutions $x = 2, x=-2$, but one of which is not a solution to the first equation.
Check: We can show that the other solution is incorrect by substituting it back into the equation and noting that $\sqrt{(x^3+5)^2} = x^3 + 5$ for $x > 0$
$$ y_2' = \frac{36}{(x^3+5)^2} - \frac{216x^3}{(x^3+5)^3}$$ $$ \frac{y_2}{x} + \sqrt{x^3{y_2}^3} = \frac{36}{(x^3+5)^2} + \frac{216x^3}{(x^3+5)^3} $$
On the other hand $\sqrt{(x^3-7)^2} = -(x^3-7)$ for $x < \sqrt[3]{7}$ and we have $$ y_1' = \frac{y_1}{x} + \sqrt{x^3{y_1}^3} = \frac{36}{(x^3-7)^2} - \frac{216x^3}{(x^3-7)^3}$$
Recap: WolframAlpha isn't right, unfortunately, and we have only one solution $$ y(x) = \frac{36x}{(x^3-7)^2}, \quad 0 < x < \sqrt[3]{7}$$