Evaluation of the value of a function at a point in a Non-linear initial value problem Given $y''y^3=1,\quad y(0)=1,\quad y'(0)=2$. I have to show that $y(1)=\sqrt{10}$. I tried to substitute $y^2=z$ but did not succeed. 

You have
$$y^{\prime} y^{\prime \prime} y^3= y^{\prime} \iff y^{\prime} y^{\prime \prime}= \frac{y^{\prime}}{y^3} \text{ (for } y^{\prime} \neq 0)$$ hence
$$\frac{1}{2}\left( y^{\prime} \right)^2 - \frac{1}{2}\left( y^{\prime}_0 \right)^2 = \frac{1}{2} \frac{1}{y_0^2} - \frac{1}{2} \frac{1}{y^2}$$ or
$$\left( y^{\prime} \right)^2-4=1-\frac{1}{y^2} \iff y^\prime = \sqrt{5-\frac{1}{y^2}}$$ as $y^\prime(0) = 2 >0$ which leads to
$$\frac{1}{5} y \sqrt{5 -\frac{1}{y^2}} - \frac{1}{5} y_0 \sqrt{5 -\frac{1}{y^2_0}} = x-x_0$$ hence
$$\frac{1}{5} \sqrt{5 y(1)^2 -1} - \frac{2}{5} = 1$$ and by solving this algebraic equation
$$y(1) = \sqrt{10}$$
