I have to prove that if $$u=t^{\lambda}y(z)$$ and $$z=\frac{x}{\sqrt{t}} \,\,,$$ then $$\frac{\partial{u}}{\partial{t}}=\frac{\partial ^{2}{u}}{\partial{x}^{2}} \Rightarrow y''_{zz}+\frac{1}{2}zy'_z-\lambda y=0$$
What would be, at least, the right direction in which I should think while solving this problem?
$\dfrac{\delta{u}}{\delta{t}} = \lambda\cdot t^{\lambda - 1}\cdot y + t^{\lambda}\cdot y_{z}'\cdot \dfrac{-x}{2}\cdot t^{-\frac{3}{2}}$, and
$\dfrac{\delta^2{u}}{\delta{x}^2} = \dfrac{\delta}{\delta{x}}\left(t^{\lambda}\cdot y_{z}'\cdot t^{-\frac{1}{2}}\right) = t^{\lambda}\cdot t^{-\frac{1}{2}}\cdot y_{z}''\cdot t^{-\frac{1}{2}}$. Thus:
$\dfrac{\lambda\cdot y\cdot t^{\lambda}}{t} - \dfrac{t^{\lambda}\cdot z\cdot y_{z}'}{2t} = \dfrac{t^{\lambda}\cdot y_{z}''}{t}$, and the answer follows from this equation by dividing both sides by $\dfrac{t^{\lambda}}{t}$