I have given it a go and found one part of the answer, which is obtained by simply using multipliers $1,1,1$ and hence $u = x+y+z$.
In the second part, I use multipliers $x,z,y$ in order, and hence the denominator reduces to $0$, therefore the numerator is equal to $0$, thus
$x \Bbb dx + z \Bbb dy + y \Bbb dz = 0 \\ x \Bbb dx + \Bbb d(zy+yz)=0 \\ x \Bbb dx + \Bbb d(2yz)=0 \\ x \Bbb dx + 2 \Bbb d(yz)=0 .$
On integration we get $\frac {x^2} 2 +2yz=v$, but in my book the answer is $\frac {x^2} 2 +yz=v$.
Where am I going wrong?
Note that $z \Bbb d y + y \Bbb d z = \Bbb d(yz)$, not $\Bbb d(zy + yz)$. This solves your problem.