If $f \in C^\infty(\mathbb{R}^2)$ with $f(0,0)=\frac{\partial f} {\partial x}(0,0)=\frac{\partial f} {\partial y}(0,0)=0$. Defining $g(t,u)=\frac{f(t,ut)}{t}$ for $t \neq 0$ and $g(0,u)=0$. I want to show that $g \in C^\infty(\mathbb{R}^2)$. Using Taylor's theorem with rest I know that there is $g_{11}, g_{12} \text{, } g_{22} \in C^\infty(\mathbb{R}^2) $ such that $$g(t,u)=tg_{11}(t,tu)+tug_{12}(t,tu)+tug_{22}(t,tu)$$
Hence it is clear that $g$ is smooth for $t \neq 0$. How to prove that at $t = 0$ $g $ is smooth? I tried to show that all partial derivatives exist and are continuous by limit, but it doesn't look promising.
The result you want follows by applying the following to $h(x,y) = f(x,xy)$.
Lemma: Suppose $h \in C^{\infty}(\mathbb R^2)$. Define $$ g(x,y) = \cases{\displaystyle\frac{h(x,y) - h(0,y)}x & if $x \ne 0$ \\ \displaystyle\frac{\partial h}{\partial x}(0,y) & if $x = 0$.} $$ Then $g \in C^{\infty}(\mathbb R^2)$.
Proof: Since $\frac{\partial h}{\partial x}$ is $C^\infty$, it can be shown that $g$ is $C^\infty$ using the formula: $$ g(x,y) = \int_0^1 \frac{\partial h}{\partial x}(tx,y) \, dt $$ and passing derivatives under the integral.
To show the formula, note first it is trivial if $x = 0$. Otherwise, apply the change of variable $s = tx$ to obtain $$ \int_0^1 \frac{\partial h}{\partial x}(tx,y) \, dt = \frac1x \int_0^x \frac{\partial h}{\partial x}(s,y) \, ds $$ and then apply the fundamental theorem of calculus.