Proof of Products and Quotients of $C^k$ Real-Valued Maps on Euclidean Open Sets Are $C^k$

We prove both parts simultaneously by induction on $k$.

Base case ($k=1$). Fix $x\in U$ and $i\in\{1,\dots,n\}$.

(i) By the definition of the partial derivative, $\dfrac{\partial(fg)}{\partial x_i}(x)$ is the limit as $h\to 0$ of $\dfrac{f(x+he_i)\,g(x+he_i)-f(x)\,g(x)}{h}$. We factor the numerator as

$$f(x+he_i)\,g(x+he_i)-f(x)\,g(x) =\bigl[f(x+he_i)-f(x)\bigr]g(x+he_i)+f(x)\bigl[g(x+he_i)-g(x)\bigr].$$

Since $f$ and $g$ are $C^1$, both partial derivatives $\dfrac{\partial f}{\partial x_i}(x)$ and $\dfrac{\partial g}{\partial x_i}(x)$ exist and $g$ is continuous. Dividing by $h$ and taking $h\to 0$: the first term tends to $\dfrac{\partial f}{\partial x_i}(x)\cdot g(x)$ (since $g(x+he_i)\to g(x)$ by continuity of $g$), and the second term tends to $f(x)\cdot\dfrac{\partial g}{\partial x_i}(x)$. Therefore

$$\frac{\partial(fg)}{\partial x_i}(x)=\frac{\partial f}{\partial x_i}(x)\,g(x)+f(x)\,\frac{\partial g}{\partial x_i}(x).$$

The right-hand side is a sum of products of continuous functions (namely $\dfrac{\partial f}{\partial x_i}$, $g$, $f$, $\dfrac{\partial g}{\partial x_i}$, all continuous by $C^1$), hence continuous by continuity of sums and products of continuous real-valued functions. As $i$ was arbitrary, $fg$ is $C^1$, proving (i).

(ii) With $g(x)\ne 0$ for all $x\in U$, the difference quotient for $f/g$ at $x$ in direction $e_i$ equals

$$\frac{\dfrac{f(x+he_i)}{g(x+he_i)}-\dfrac{f(x)}{g(x)}}{h} =\frac{\bigl[f(x+he_i)-f(x)\bigr]g(x)-f(x)\bigl[g(x+he_i)-g(x)\bigr]}{h\cdot g(x+he_i)\cdot g(x)}.$$

Dividing the numerator by $h$ and taking $h\to 0$: the two fractions tend to $\dfrac{\partial f}{\partial x_i}(x)$ and $\dfrac{\partial g}{\partial x_i}(x)$ respectively, and $g(x+he_i)\to g(x)\ne 0$ by continuity. Therefore

$$\frac{\partial(f/g)}{\partial x_i}(x)=\frac{\dfrac{\partial f}{\partial x_i}(x)\,g(x)-f(x)\,\dfrac{\partial g}{\partial x_i}(x)}{g(x)^2}.$$

The numerator and denominator $g^2$ are continuous (since $g$, $\dfrac{\partial f}{\partial x_i}$, $\dfrac{\partial g}{\partial x_i}$ are continuous and $g^2\ne 0$), so $\dfrac{\partial(f/g)}{\partial x_i}$ is continuous by continuity of sums and products. As $i$ was arbitrary, $f/g$ is $C^1$, proving (ii).

Inductive step. Assume both (i) and (ii) hold for $k-1$ in place of $k$; let $f,g:U\to\mathbb{R}$ be $C^k$ with $k\ge 2$.

(i) For each $i\in\{1,\dots,n\}$, the base-case computation (which is a pointwise limit argument valid for any $C^1$ functions) gives $\dfrac{\partial(fg)}{\partial x_i}=\dfrac{\partial f}{\partial x_i}\cdot g+f\cdot\dfrac{\partial g}{\partial x_i}$. Since $f$ and $g$ are $C^k$, the partial derivatives $\dfrac{\partial f}{\partial x_i}$ and $\dfrac{\partial g}{\partial x_i}$ are $C^{k-1}$, and $f$, $g$ themselves are $C^{k-1}$. By the inductive hypothesis (i), both products $\dfrac{\partial f}{\partial x_i}\cdot g$ and $f\cdot\dfrac{\partial g}{\partial x_i}$ are $C^{k-1}$; their sum is also $C^{k-1}$. As $i$ was arbitrary, $fg$ is $C^k$.

(ii) Similarly, the base-case formula $\dfrac{\partial(f/g)}{\partial x_i}=\bigl[\dfrac{\partial f}{\partial x_i}\,g-f\,\dfrac{\partial g}{\partial x_i}\bigr]\big/g^2$ holds. The numerator is $C^{k-1}$ by (i) applied inductively, and $g^2$ is $C^{k-1}$ by (i) inductively, with $g^2>0$ on $U$. By the inductive hypothesis (ii), the quotient $\dfrac{\partial(f/g)}{\partial x_i}$ is $C^{k-1}$. As $i$ was arbitrary, $f/g$ is $C^k$.

By induction, both (i) and (ii) hold for all natural numbers $k\ge 1$.