Let $ f : \mathbb{R}^3 \rightarrow \mathbb{R}^3$ defined by $f(v)=\lambda v$ where $\lambda >0$ is constant. Consider a surface $S \subseteq \mathbb{R}^3$ and $S'=f(S)$. Find a relation between $K$ of $S$ and $K'$ of $S'$ where $K$ is the Gaussian curvature.
$\textbf{My attempt : }$ Let a parametrization of $S$, $\varphi(u,v) : U_0 \subseteq \mathbb{R}^2 \rightarrow S$. We have a parametrization of $S'$ defined by : $\phi = f \circ \varphi : U_0 \rightarrow S'$ (because $f$ is a diffeomorphism).
So :
$$\phi(u,v) = (f\circ \varphi) (u,v) = \lambda \varphi (u,v)$$
$$ \implies \phi_u = \lambda \varphi_u \hspace{0.5cm} \text{and} \hspace{0.5cm} \phi_v=\lambda \varphi_v $$ $$ \implies \phi_{uu} = \lambda \varphi_{uu} \hspace{0.5cm} \text{and} \hspace{0.5cm} \phi_{vv}=\lambda \varphi_{vv} \hspace{0.5cm} \text{and} \hspace{0.5cm} \phi_{uv}=\lambda \varphi_{uv} $$
Finally :
$E' = \langle \phi_u , \phi_u \rangle = \lambda^2 \langle \varphi_u , \varphi_u \rangle = \lambda^2 E$ and $L' = \langle \phi_{uu} , n \rangle = \lambda \langle \varphi_{uu} , n \rangle = \lambda L$
Analogous :
$F'=\lambda^2 F, G'= \lambda^2 G,M'=\lambda M,N' =\lambda N$
$\implies K' = \dfrac{L'N'-(M')^2}{E'G' - (F')^2} = \dfrac{\lambda^2(LN-M^2)}{\lambda^4(EG-F^2)}=\dfrac{1}{\lambda^2}K $
It´s right? My question is what would happen if I took an arbitrary parameterization of $S'$?
No need to use ugly coordinate formulas. If $N$ is a normal Gauss field for $S$, then a normal Gauss field $\widetilde{N}$ for $\lambda S$ is given by $\widetilde{N}(\lambda p) = N(p)$. The chain rule then gives that ${\rm d}\widetilde{N}_{\lambda p} = (1/\lambda){\rm d}N_p$. It follows that $\widetilde{K}(\lambda p)=K(p)/\lambda^2$ and $\widetilde{H}(\lambda p)=H(p)/\lambda$.