Need help with a first order differential equation questions

How do i solve the ODEs:
a) dy/dx = (sinx)/y
b) x*(dy/dx) = y*lnx

the answers are
a) cuberoot(C - 3*cosx)
b) (lnx)^2 = 2*lny + C

I dont know how to get these answers, show all steps thanks

Please ask one question at a time in future, makes it more manageable to answer.

1.
Solve this differential equation by separating the variables then integrating:
dy / dx = sinx / y²
y² dy = sinx dx
∫ y² dy = ∫ sinx dx
y³ / 3 = -cosx + C
y³ / 3 = C - cosx
y³ = C - 3cosx
y = ∛(C - 3cosx)

2.
Solve this differential equation by separating the variables then integrating:
x(dy / dx) = ylnx
dy / y = lnx / x dx
∫ 1 / y dy = ∫ lnx / x dx

Integrate the expression on the right by substitution:
∫ lnx / x dx
Let u = lnx,
du / dx = 1 / x
du = dx / x
∫ lnx / x dx = ∫ u du
∫ lnx / x dx = u² / 2 + C
Since u = lnx,
∫ lnx / x dx = (lnx)² / 2 + C

Find the solution to the differential equation using the above result:
∫ 1 / y dy = ∫ lnx / x dx
ln|y| = (lnx)² / 2 + C
ln|y| = C + (lnx)² / 2
y = ℮^[C + (lnx)² / 2]
y = ℮ᶜ℮^[(lnx)² / 2]
y = C℮^[(lnx)² / 2]

The answer to a) is wrong. The correct answer is y = sqrt(C-2cosx)

ydy = sinx dx. Integrate both sides: (1/2) y^2 = C-cosx


Use the same method for b)