Why does the definition of MRS follow from the implicit function theorem? TRAIN OF THOUGHT 1: 

From what I understand, $MRS$ is calculated as 

$$dU = U_x dx +  U_y dy =0$$ 
which by rearrangement yields 
$$\frac{dy}{dx}= -\frac{U_x}{U_y}$$

So suppose I have $$U(x,y) = \ln x +\ln y$$ Then 
$$ \frac{dy}{dx}= -\frac{1/x}{1/y} = -\frac{y}{x}$$

Okay. So I have a function $y$ in terms of $x$.

TRAIN OF THOUGHT 2: 

Now consider my $U(x,y)$ again. Let $$\mathbf{a} = \begin{bmatrix}
1\\
1
\end{bmatrix}$$ 
and $$U(\mathbf{a})=0$$
We have $$DU(x,y) = \begin{bmatrix}
\frac{1}{x} &amp; \frac{1}{y}
\end{bmatrix} $$
and $$\frac{\partial U}{\partial y} (\mathbf{a}) = \begin{bmatrix}\frac{1}{y}\end{bmatrix}= 1$$ which is nonsingular since $\det(1) = 1$ and so by the Implicit Function Theorem, $$U = 0$$ defines $y$ implicitly as a function of $x$ in a neighborhood of $\mathbf{a}$.  

My Question:

How are these two trains of thought connected? The first is stated in terms of differentials. But the second is not. So I am confused why the definition of $MRS$ follows from the implicit function theorem.
It's actually pretty straight forward. The implicit function theorem for two variables is given as follows (as long as some regularity conditions hold):

For $F(x, y) = 0$,

$
\frac{dy}{dx} = -\frac{\partial F / \partial x}{\partial F / \partial y}
$

In the case of MRS, we want the marginal change in $x$ associated with a marginal change in $y$ required to maintain a certain level of utility, $c$, such as (conveniently) $c=0$. So, starting with

$U = U(x, y) = 0$,

we have

$\frac{dy}{dx} = -\frac{U_x}{U_y}$

Note that $c=0$ is just a simplification for exposition. For a general $c$, you can just subtract it from either side of the equation and you get the same result since $c$ disappears in the derivative.