Differential equation question calculus

Can someone please help me through this problem with steps , the condition of y(3) = -2 is confusing me like I don't know where to plug it in.
What is the solution to the differential equation dy / dx = 3x^2 / y with the initial condition y(3) = -2
Thanks so much

Find the general solution by separating the variables then integrating:
dy / dx = 3x² / y
y dy = 3x² dx
∫ y dy = ∫ 3x² dx
y² / 2 = x³ + C
y² = 2x³ + C
y = ±√(2x³ + C)

Find the particular solution by solving for the constant:
When x = 3, y = -2
±√(54 + C) = -2
54 + C = 4
C = -50
y = ±√(2x³ - 50)
y = -√(2x³ - 50)

This is a first-order differential equation with initial condition y(3) = -2 that can be solved by separation method:

dy/dx = (3x^2)/y
ydy = (3x^2)dx

Now integrating both sides:

(1/2)y^2 = (x^3) + Constant

Solving for y:

y^2 = 2(x^3) + Constant <---Notice it's not 2*Constant because Constant is arbitrary
y = (+/-)(2(x^3) + Constant)^0.5

Now using initial condition of y(3) = -2 to solve for Constant:

y(3) = -2 = (+/-)(2(3^3) + Constant)^0.5
y(3) = -2 = (+/-)(54 + Constant)^0.5
((+/-)2)^2 = 54 + Constant
4 = 54 + Constant
Constant = -50

Substituting Constant back into solution:

y = (+/-)(2(x^3) + Constant)^0.5
y = (+/-)(2(x^3) - 50)^0.5 <----Answer, where x^3 >= 25