The left side is the derivative of [tex](y')^2[/tex]:
[tex]\left((y')^2\right)'=2y'y''[/tex]
So we can integrate both sides of
[tex]\left((y')^2\right)'=1\implies (y')^2=x+C\implies y'=\pm\sqrt{x+C}[/tex]
Then integrate again to solve for [tex]y[/tex]:
[tex]y=\pm\dfrac23(x+C_1)^{3/2}+C_2[/tex]
With the given initial conditions, we find
[tex]y(0)=2\implies 2=\pm\dfrac23{C_1}^{3/2}+C_2[/tex]
[tex]y'(0)=1\implies 1=\pm\sqrt{C_1}[/tex]
The second equation says [tex]C_1[/tex] is either 1 or -1, but in the latter case, we would get [tex](-1)^{3/2}=\sqrt{-1}[/tex] in the first equation, which is undefined over the real numbers, so [tex]C_1=1[/tex].
So there are two candidate solutions,
[tex]y_1=\dfrac23(x+1)^{3/2}+\dfrac43[/tex]
[tex]y_2=-\dfrac23(x+1)^{3/2}+\dfrac83[/tex]
However, the second equation doesn't satisfy the initial value of the derivative, since [tex]{y_2}'(0)=-1\neq1[/tex]. So the solution is
[tex]\boxed{y(x)=\dfrac23(x+1)^{3/2}+\dfrac43}[/tex]
