Solve the differential equation 6y dy/dx = 6x*(1+x^2)^(1/2)*(1+2y^2)^(1/2) subject to the initial condition...

Initially I might start out like this:

6y dy = 6x√(1+x²) • √(1+2y²) dx now get the y-factor on the right over to the left by dividing ....

6y dy/√(1+2y²) = 6x√(1+x²) dx Now try to integrate both sides. (Those 6's might cancel or you may need them in the integration ... let's find out)

∫ 6y dy/√(1+2y²) = ∫ 6x√(1+x²) dx

3√(2y² + 1) = 2(x²+1)^3/2 + constant

Now apply the initial condition y(0) = 0 ... and so it goes ... for separable DEs.

Are you sure you've got the right answer?
Because I get

6y dy/dx = 6x*(1+x^2)^(1/2)*(1+2y^2)^(1/2)
6y/(1+2y^2)^(1/2) dy/dx = 6x*(1+x^2)^(1/2)

int[6y/(1+2y^2)^(1/2) dy] = int[6x*(1+x^2)^(1/2) dx]
3(1+2y^2)^(1/2) = 2(1+x^2)^(3/2) + C

now the initial condition comes in - set x,y to 0

3(1+2(0)^2)^(1/2) = 2(1+(0)^2)^(3/2) + C
3 = 2 + C
C = 1

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