I think there might be a mistake in this attempted proof of Lagrange Multipliers, but I'm not sure where. Halfway through I set the total derivative of the objective function to $0$. I believe the usual proof asserts the stationary point occurs where the gradient of the objective function minus it's component along the gradient of the constraint function is 0, and not the full differential yet I'm getting the same results. Is the "Suspicious" step a problem below?
Suppose your constraint is $g(x,y,z)=0$ and your objective function is $f(x,y,z)$.
Then if we are any where on the level curve, $0=dg = \nabla g \cdot \vec{ds}$. We need a $\vec{ds}$ perpendicular to $\nabla g$.
For an arbitrary vector $\vec{u} $ we get this if $ \vec{ds} = \vec{u} \times \nabla g$.
SUSPICIOUS STEP? Require $df=\nabla f \cdot \vec{ds}=0$ to get an extremum. $0=df=\nabla f \cdot (\vec{u} \times \nabla g)$.
This is the vector triple product and can be rewritten as $0=\vec{u} \cdot (\nabla g \times \nabla f)$
For this to be true for arbitrary $\vec{u}$ we need $\nabla g \times \nabla f = \vec{0}$. This implies the gradients are parallel, giving us the same result as usual Lagrange Multipliers proofs which usually requires taking a derivative with respect to the multiplier.
This implies $0=-\nabla g \times (\nabla f\times \nabla g)= -( \nabla f (\nabla g)^2 - \nabla g(\nabla f \cdot \nabla g) )\implies \nabla f = \frac{\nabla f \cdot \nabla g}{\nabla g \cdot \nabla g} \nabla g=0 $ this being the typical Lagrange Multiplier Criterion.
NOTES ON NOTATION.
Here's a guide to my notation.
For $f(x,y,z)$ a differentiable function $R^3 \to R$ in cartesian coordinate system.
$\nabla f = \frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}+\frac{\partial f}{\partial z}\hat{k}$
$\vec{ds}$ is a differential line element.
In cartesian coordinates its $\vec{ds} = dx\hat{i} + dy\hat{j} + dz\hat{k}$ or $\vec{ds}= dr \hat{r} + r d\theta \hat{\theta} + dz \hat{k} $ in cylindrical coordinates.
Consequently $\nabla f \cdot \vec{ds} = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z}dz$, the total derivative of $f$.
I can't say if this is a valid proof or not because I cannot understand what your notation is supposed to mean:
In the 3rd paragraph, you write $dg = \nabla g\cdot \vec{ds} =0$. What is this supposed to mean? The level set $L$ you want to optimize $f$ over is given by the constraint $g(x,y,z)=0$, so I think you want $dg$ to be the derivative of the restriction of $g$ to $L$, which is going to be built out of $Dg$, the total derivative of $g$, in one way or another, and Dg is a function taking values in the space of linear maps $\text{Hom}(\mathbb R^3,\mathbb R)$.
The symbols $\nabla g\cdot \vec{ds}$ don't give much assistance here, because, whatever $\vec{ds}$ is supposed to be, given you take its dot product with $\nabla g$, it would appear to be a vector or perhaps a vector-valued function. Regardless, $\nabla g\cdot \vec{ds}$ will then be a scalar valued function, while $dg$ should certainly not be scalar valued.
You also write "to get an extremum" for $f$ you require $df = \nabla f\cdot \vec{ds} =0$ and not only does this has the same problems as above in that it does not make sense, it also leaves a logical gap: Lagrange multipliers give a necessary condition for an extremum, so one must start with a point $x_0$ where $f$ has a constrained extremum and show how the condition relating $\nabla f$ and $\nabla g$ can be deduced. Why, if $f$ has a local extremum at $x_0$, must it follow that $df=\nabla f\cdot \vec{ds}=0$ (again, whatever that is supposed to mean)?
Update: You write that $\vec{ds} = dx \hat{i}+ dy\hat{j} + dz\hat{k}$. Assuming $x,y,z$ are, as usual, coordinate functions on $\mathbb R^3$, with total derivative $dx,dy$ and $dz$, then this definition give $\vec{ds} = I_3$ the identity linear map from $\mathbb R^3$ to itself. That does indeed make $dg = \nabla g\cdot \vec{ds}$ into a valid formula, though I think it is much clearer to say that $\nabla g_x$ , the value of $\nabla g$ at $x$, is uniquely determined by the condition that $df_x(v) = \nabla g_x\cdot v$ for all $v \in\mathbb R^3$.
That said, I think this does mean I can interpret what you wrote more readily, and my objection above appears to stand: You say "require $df = \nabla f \cdot \vec{ds} =0$ to get an extremum". Apart from the fact that $df$ is not (necessarily) zero, this is simply assuming a version of the Lagrange multipliers theorem in order to prove it.
To prove that the Lagrange multiplier condition gives a necessary condition for a point $x_0 \in U$ to be a local minimum, you must deduce that $\nabla g_{x_0}$ is parallel to $\nabla f_{x_0}$ from the assumption that $f(x_0)\leq f(x)$ for all $x \in L = \{x: g(x)=0\}$ sufficiently close to $x_0$ (or $f(x_0)\geq f(x)$ for all $x\in L$ sufficently close to $x_0$. As far as I can see, you do not use the local extremum condition anywhere in your argument.