Is the differential forms perspective on $dx$ incompatible with the technique of implicit differentiation? Suppose $$x^2 + y^2 = 5^2.$$ We're trying to find $dy/dx$ at $(3,4).$
Applying $d$ to both sides: $$2x dx + 2y dy = 0$$
Or in other words:
$$2x dx + 2y dy = 0dx + 0dy$$
Since the covectors $dx_p$ and $dy_p$ form a basis for the cotangent space at any $p \in \mathbb{R}^2$, hence $2x = 0$ and $2y = 0.$ Hence $x = 0$ and $y = 0$. Ergo $0^2 + 0^2 = 5^2$, a contradiction.

Question. Does this mean that the differential forms perspective on $dx$ is incompatible with the technique of implicit differentiation?
If not, why not?
If so, what definition of $dx$ can be used to avoid this issue?

 A: What you're missing is that this computation does not take place in $\mathbb{R}^2$: it takes place in the submanifold $M$ of $\mathbb{R}^2$ defined by the equation $x^2+y^2=5^2$!  We can consider $x$ and $y$ as smooth functions on $M$, and these smooth functions satisfy $x^2+y^2=5^2$ at every point of $M$.  So, we can differentiate and deduce that $$2xdx+2ydy=0$$ at every point of $M$.  This is not a contradiction, because $dx$ and $dy$ here are sections of the cotangent bundle of $M$, not the cotangent bundle of $\mathbb{R}^2$ (they are the pullback of the usual $dx$ and $dy$ on $\mathbb{R}^2$ to $M$).
Note that on $\mathbb{R}^2$, you cannot differentiate the equation $x^2+y^2=5^2$ since this is not actually an equation of functions, it is just an equation that is true at some points of $\mathbb{R}^2$.  Even at a point where the equation is true, you cannot expect the differential of $x^2+y^2$ to be equal to the differential of $5^2$, since there are nearby points where the two functions are not equal.  But we don't have this problem when we restrict to $M$, since $x^2+y^2$ and $5^2$ really are the same function on $M$.
(The elementary calculus justification for this implicit differentiation is to say we are not differentiating on $\mathbb{R}^2$ but that instead we are considering $x$ and $y$ as functions of some other variable $t$ via a parametrization of our curve, and then differentiating with respect to $t$.  (Or, if we're finding $dy/dx$ as you suggest, we're treating $y$ as a function of $x$ locally on the curve so the parameter is $x$ itself.)  This is in fact exactly the same as the manifold approach when you unwrap all the definitions, since the differential on $M$ is defined in terms of differentiating in a local parametrization.)
