Using the chain rule to find second partial derivatives of $f(\alpha x, \alpha y)$ Consider the function $f( \alpha x, \alpha y)$ for some $\alpha \in \mathbb{R}$. We wish to compute the second partial derivative of $f$ with respect to $x$, that is $f_{xx}$. Is the following method correct?
First we define a change of variables : $$\bar{x} =\alpha x $$ $$\bar{y} = \alpha y $$
Then we use the chain rule to find the first partial derivative of $f$ with respect to $x$: $$f_{x}=\alpha f_{\bar{x}}$$
Now we wish to compute the second partial derivative, $$f_{xx}=\frac{\partial f}{\partial x}(f_{x})=\frac{\partial f}{\partial x}(\alpha f_{\bar{x}})=\alpha \frac{\partial f}{\partial \bar{x}}(f_x)=\alpha \frac{\partial f}{\partial \bar{x}}(\alpha f_\bar{x})=\alpha^2f_{\bar{x}\bar{x}} $$
Thus we have that $f_{xx} = \alpha^2 f_{\bar{x}\bar{x}}$. Does this work?
Edit: I have been informed my computation is incorrect, what is the correct way?
 A: It might be helpful to perform the chain rule calculation in a slightly more general setting. We consider functions
\begin{align*}
&f:\mathbb{R^2}\to\mathbb{R}&\ \ g:\mathbb{R}\to\mathbb{R}\quad&\qquad h:\mathbb{R}\to\mathbb{R}\\
&(x,y)\to f(x,y)&x\to \alpha x\qquad &\qquad y\to \alpha y\\
\end{align*}
We obtain by applying chain rule and product rule
\begin{align*}
\color{blue}{\frac{\partial^2}{\partial x^2}f\left(g(x),h(y)\right)}
&=\frac{\partial}{\partial x}\left(f^{\prime}\left(g(x),h(y)\right)g^{\prime}(x)\right)\\
&\,\,\color{blue}{=f^{\prime\prime}\left(g(x),h(y)\right)\left(g^{\prime}(x)\right)^2+f^{\prime}\left(g(x),h(y)\right)g^{\prime\prime}(x)}\tag{1}
\end{align*}
Since
\begin{align*}
g(x)&=\alpha x\qquad\qquad h(y)=\alpha y\\
g^{\prime}(x)&=\alpha\\
g^{\prime\prime}(x)&=0
\end{align*}
we get from (1) noting that $g^{\prime\prime}(x)=0$
\begin{align*}
\color{blue}{\frac{\partial^2}{\partial x^2}f\left(\alpha x,\alpha y\right)=f^{\prime\prime}\left(\alpha x,\alpha y\right)\alpha^2}
\end{align*}
