[tex]y=\sqrt{x(x+2)}=\sqrt x\sqrt{x+2}=x^{1/2}(x+2)^{1/2}[/tex]
Take the logarithm of both sides.
[tex]\ln y=\ln(x^{1/2}(x+2)^{1/2})=\dfrac12\ln x+\dfrac12\ln(x+2)[/tex]
In the last equality, we used the fact that [tex]\ln(ab)=\ln a+\ln b[/tex] and [tex]\ln a^c=c\ln a[/tex].
Differentiating both sides with respect to [tex]x[/tex], we have
[tex]\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1{2x}+\dfrac1{2(x+2)}[/tex]
where the left hand side occurs by the chain rule (recall that [tex]y[/tex] is a function of [tex]x[/tex]).
The solution then follows from the steps in your attachment, which is just a matter of algebraic manipulation.
