BY taking 1st and 2nd equations 1st solution is $$ x=y+c_{0}$$

but i am getting two different solutions slightly differing :

Solution 1 : using lagrange multiplier 1,1,0 we get

$$ frac{dx+dy}{2z}=frac{(3z+1)dz}{x+y}$$

$$ frac {(x+y)d(x+y)}{2z}= {(3z+1)(dz)}$$

$$ frac {d(x+y)^{2}}{2} = (2z)(3z+1)(dz) =(6z^{2}+2z)dz$$

$$ frac {(x+y)^2}{2} = {2z^3}+(z^2)+c_{1} $$

$$ (x+y)^{2}= 4z^{3}+2z^{2}+c_{1′}$$

Solution 2 : taking x,y,-z as multipliers

$$ frac {xdx+ydy-(zdz)(3z+1)}{xz+yz-z(x+y)} =frac {xdx+ydy-(zdz)(3z+1)}{0}$$

$$ xdx+ydy= (3z+1)(zdz)=(3z^{2}+z)dz$$

$$ frac{x^2}{2} +frac{y^2}{2}= z^{3}+ frac{z^2}{2}+c_{2} $$

$$ x^{2}+y^{2}=2z^{3}+z^{2}+c_{2′}$$

$$ 2(x^{2}+y^{2})=4z^{3}+2z^{2}+c_{2”}$$

here in the first solution the term of $$xy$$ is extra .. i believe i have done mistake in my first solution but i am unable to find out where?

solution 1 is my solution and solution 2 is the solution from the book