square?

$\begin{aligned} \sqrt {x^3} &= - 4x \\ x ^\frac 32 +4x& =0 \\ x (x ^\frac 12 +4)&=0 \\ \implies x&=0 & \small \color{#3D99F6} {x ^\frac 1 2+4=0 \text { has no real root.}} \end{aligned}$

Therefore, there is only $\boxed {1}$ solution.

$\sqrt{x^3} = -4x\\ x^3=16x^2\\ x^3-16x^2=0\\ x^2(x-16)=0\\ x=0,\;x=16$

Note that we squared the original equation in the beginning. This implies that we might have introduced extra roots which do not satisfy the original equation. Let us check:

$x=0,\;\sqrt{x^3} = \sqrt{0^3} = 0,\;-4x=-4(0)=0\\ x=16,\;\sqrt{x^3}=\sqrt{16^3}=64,\;-4x=-4(16)=-64$

From here, we can see that $x=16$ is an extra root. The only actual solution to the equation is $x=0$

There is only $\boxed{1}$ solution

$\sqrt [ 2 ]{ { x }^{ 3 } } \ge 0\quad \Longrightarrow x\ge 0$ (1)

AND!!!! $-4x\ge 0\quad \Longrightarrow \quad x\le 0$ (2)

so, from (1),(2) $0\le x\le 0$ $\Longrightarrow \quad x=0$

and so it has only 1 solution x=0