$u=x^2+y^2\text{ and }x=s+3t,y=2s-t,\text{ then }\frac{d^{2}u}{d{s}^2}$ is

Given $u=x^2+y^2,x=s+3t,y=2s-t$
$\begin{array}{l}\frac{\mathrm{dx}}{\mathrm{ds}}=1,\frac{\mathrm{dy}}{\mathrm{ds}}=2;\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{ds}}^2}=0,\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{ds}}^2}=0\\ \mathrm{u}={\mathrm{x}}^2+{\mathrm{y}}^2\Rightarrow \frac{\mathrm{du}}{\mathrm{ds}}=2\mathrm{x}\frac{\mathrm{dx}}{\mathrm{ds}}+2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{ds}}\end{array}$
$\begin{array}{r}=\frac{d^{2}u}{d{s}^2}=2{\left(\frac{dx}{ds}\right)}^2+2x\frac{d^{2}x}{d{s}^2}+2{\left(\frac{dy}{ds}\right)}^2+2y\left(\frac{d^{2}y}{d{s}^2}\right)\\ \Rightarrow \frac{d^{2}u}{d{s}^2}=2(1{)}^2+2x(0)+2(2{)}^2+2y\left(0\right)=2+8=10\end{array}$