A $C^1$ embedded hypersurface $M\subseteq {\mathbb{R}}^n$ is locally carried by a $C^1$ coordinate diffeomorphism to an open subset of ${\mathbb{R}}^{n-1}\times \{0\}$. Equivalently, near each of its points it has the form $F^{-1}(0)$ for a $C^1$ function with nonzero gradient there.

The Lagrange multiplier theorem states that if a differentiable function $g$ has a local extremum on such a hypersurface at $x_0$, then $dg(x_0)$ annihilates the tangent space. In a regular local level-set description this isfor some real $\lambda$. The regularity condition $\nabla F(x_0)\ne 0$ is essential; it is what identifies the normal line and permits the local constraint chart.

Differentiability at $x_0$ means that a linear map $L:{\mathbb{R}}^n\to {\mathbb{R}}^m$ exists such that $f(x_0+h)=f(x_0)+Lh+o(\Vert h\Vert )$. Since finite-dimensional linear maps are bounded, the right side tends to $f(x_0)$ as $h\to 0$, proving continuity.

For the specified scalar function, $0\le f(x,y)\le x^2\le x^2+y^2$. Dividing by $\sqrt{x^2+y^2}$ gives a bound tending to zero at the origin. Consequently $\boxed{df(0,0)=0}$ and the function is differentiable there.

For the final product assertion, write $f(x_0+h)=Lh+r(h)$, where $r(h)=o(\Vert h\Vert )$, and $g(x_0+h)=g(x_0)+s(h)$ with $s(h)=o(1)$. Since $f(x_0)=0$,Cauchy–Schwarz and boundedness of $g$ near $x_0$ justify the last estimate. Thus $\boxed{dF(x_0)h=⟨df(x_0)h,g(x_0)⟩}$; differentiability of $g$ is unnecessary because $f(x_0)=0$.

A direct derivative bound by $1/x$ would not be integrable, so the earlier theorem cannot be applied that way. Instead, for $t>0$ make the legitimate substitution $s=tx$ in the absolutely convergent integral:The tail integral has continuous integrand away from zero and converges absolutely. The fundamental theorem of calculus therefore gives derivative $-\cos t/t^2$ with respect to its lower endpoint. The product rule yieldsThis establishes differentiability carefully without an invalid domination argument.