dy\dx=-(3x^2y+y^2)\2x^3+3xy

Question

shoe steps

Answer 1

Remeber the formulae d/dx(uv) = uv'+vu'

d/dx(u/v) = (vu'-uv')/v^2

y' = -(3x^2y+y^2)/(2x^3+3xy)

Apply derivative on each side with respect of x.

Answer 2

dy/dx = - (3x²y + y²)/(2x³ + 3xy)
The rule for implicit differentiation in two variables is
dy/dx = - Fx/Fy

where F(x, y) = 0 is the solution, and Fx, Fy are partial derivatives
as the ratio is (3x²y + y²)/(2x³ + 3xy)

Fx is proportional to 3x²y + y² and
Fy to 2x³ + 3xy

I mean that num and den can be multiplied for a factor such that integrating
the num wrt x and the den wrt y the result is the same function

if I multiply Fx and Fy by y I get the same ratio Fx/Fy
and
∫(3x²y² + y³) dx = x³y² + xy³ + f(y) + C
∫(2x³y + 3xy²) dy = x³y² + xy³ + g(x) + C, g(x) and f(y) are arbitrary functions
we set f(y) = 0, g(x) = 0 to make the two integral identical
F(x,y) = x³y² + xy³ + C = 0
F(1,-2) = 0, C = 4

x³y² + xy³ + 4 = 0

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