Let [tex]x=t[/tex], so that
[tex]x^2=2y\implies t^2=2y\implies y=\dfrac{t^2}2[/tex]
[tex]3z=xy\implies3z=\dfrac{t^3}2\impiles z=\dfrac{t^3}6[/tex]
Then the length of the path is
[tex]\displaystyle\int_{t=0}^{t=4}\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2+\left(\frac{\mathrm dz}{\mathrm dt}\right)^2}\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^4\sqrt{1+t^2+\frac{t^4}4}\,\mathrm dt[/tex]
[tex]=\displaystyle\frac12\int_0^4\sqrt{4+4t^2+t^4}\,\mathrm dt[/tex]
[tex]=\displaystyle\frac12\int_0^4\sqrt{(t^2+2)^2}\,\mathrm dt[/tex]
[tex]=\displaystyle\frac12\int_0^4(t^2+2)\,\mathrm dt[/tex]
[tex]=\dfrac{44}3[/tex]
