Find the dy/dx?

Question

ln x + ln(y^2) = 3

Answer 1

(1/x) + (1/y^2)(2y)y' = 0
(2 / y)y' = -(1/x)
y' = (-1/x)(y/2)
y' = -y / 2x

Answer 2

i propose to pass from the logarithm to power
remember lna+ln b= ln ab
first remark that x must be positive else no log

so lnxy^2 = lne^3

xy^2 = e^3

y = e^3/2 /x^1/2

dy/dx= -1/2 *e^(3/2)*x^(-3/2) or -1/2 *e^(3/2) /(x^3)^0.5

Answer 3

lnx + ln(y^2) = 3
hence

lnx + 2lny = 3

no differentiating with respect to x , we get :

1/x + (2/y)dy/dx = 0

hence

dy/dx = -y/2x