Simple proof,

By Taylor formula $f(x) = \sum\limits_{k = 0}^{n-1 } \frac{f^{(2k)}(0)}{(2k)!}x^{2k} + x^{2n }\epsilon (x)$ with $\epsilon(x)=\int_0^1 \frac{(1-t)^{2n-1}}{(2n-1)!} f^{(2n)}(tx)dt$ , $\epsilon\in \mathcal{C}^{\infty}(\mathbb{R}, \mathbb{R})$ because $f \in \mathcal{C}^{\infty}(\mathbb{R}, \mathbb{R})$ then all dérivatives of $\epsilon $ are bounded in neighborhood of 0.

$$\forall x\geq0,\quad g(x)= \sum\limits_{k = 0}^{n-1 } \frac{f^{(2k)}(0)}{(2k)!}x^{k} + x^{n }\epsilon (\sqrt x)=g_1(x)+g_2(x)$$ with $g_2(x)=x^{n }\epsilon (\sqrt x)$. To show $g^{(n)}(0)$ exists , it suffices to show $g_2^{(n)}(0)$ exists $$g_2'(x)=x^{n-1}(n\epsilon(\sqrt x)+\sqrt x \epsilon'(\sqrt x))=x^{n-1 }\epsilon_1 (\sqrt x)$$ with $\epsilon_1(x)= n\epsilon (x)+x\epsilon'(x)\to 0$ because $\epsilon '$ is bounded in neighborhood of 0. By induction $$g_2^{(k)}(x)=x^{n-k}\epsilon_k(\sqrt x),\quad \forall k\leq n $$ thus $g_2^{(n)}(x)=\epsilon_n(\sqrt x)$ and $g_2^{(n)}(0)$ exists

Basic proof,

By Taylor formula $f(x) = \sum\limits_{k = 0}^{n-1 } \frac{f^{(2k)}(0)}{(2k)!}x^{2k} + x^{2n }\epsilon (x)$ with $\epsilon(x)=\int_0^1 \frac{(1-t)^{2n-1}}{(2n-1)!} f^{(2n)}(tx)dt$ , $\epsilon\in \mathcal{C}^{\infty}(\mathbb{R}, \mathbb{R})$ because $f \in \mathcal{C}^{\infty}(\mathbb{R}, \mathbb{R})$ then all dérivatives of $\epsilon $ are bounded in neighborhood of 0.

$$\forall x\geq0,\quad g(x)= \sum\limits_{k = 0}^{n-1 } \frac{f^{(2k)}(0)}{(2k)!}x^{k} + x^{n }\epsilon (\sqrt x)=g_1(x)+g_2(x)$$ with $g_2(x)=x^{n }\epsilon (\sqrt x)$. To show $g^{(n)}(0)$ exists , it suffices to show $g_2^{(n)}(0)$ exists $$g_2'(x)=x^{n-1}(n\epsilon(\sqrt x)+\sqrt x \epsilon'(\sqrt x))=x^{n-1 }\epsilon_1 (\sqrt x)$$ with $\epsilon_1(x)= n\epsilon (x)+x\epsilon'(x)\to 0$ because $\epsilon '$ is bounded in neighborhood of 0. By induction $$g_2^{(k)}(x)=x^{n-k}\epsilon_k(\sqrt x),\quad \forall k\leq n $$ thus $g_2^{(n)}(x)=\epsilon_n(\sqrt x)$ and $g_2^{(n)}(0)$ exists