Here is perhaps another way that I have found to think about this.
We want the change in $f(x,y)$, $-\frac{f_x}{f_y}$ which I will call $df$. We can use the partial derivatives of $f$ to write a linear approximation of $f$ near some point on the level curve, $(a,b)$. This linear approximation would be: $$L(i,s) = f(a,b) + [f_x(a,b)](i-a)+[f_y(a,b)](s-b)$$
This linear approximation is very good very close (infinitesimally close) to $(a,b)$ (this fact and how to find the linear approximation/why it is written as above can be found in calculus books or through a google search). Therefore, an infinitesimally small movement along the level curve, $df$ (a first order change) can be written as $$L(i + di,s + ds) - L(i,s) =0$$ because $df =0$ (since we are on a level curve).
Some algebra will give us $[f_x(a,b)]dx+[f_y(a,b)]dy =0$ (i replaced di and ds with dx and dy since they are movements in the same directions), which is the slope of the linear approximation and thus the slope of the level curve.
Yes, there are several abuses of notation here. What is happening is you're first given a smooth function $F:\Bbb{R}^2\to\Bbb{R}$; for simplicity assume that at every point $p\in\Bbb{R}^2$, we have $\frac{\partial F}{\partial y}(p)\neq 0$. The implicit function theorem tells us that if you fix such a point $p=(a,b)$, then you can find some smooth function $\eta:I\subset\Bbb{R}\to\Bbb{R}$ such that $\eta(a)=b$ and for all $t\in I$, we have $F(t,\eta(t))=0$. So, the function $w:I\to\Bbb{R}$ defined as $w(t)=F(t,\eta(t))$ is smooth and is zero at every point; i.e is the constant zero function. So, we obviously have that $w'=0$. But now what does the chain rule tell us (note that $w$ is the composition of $F$ with the function $t\mapsto (t,\eta(t))$, so chain rule is indeed the way to go)? It tells us for each $t\in I$,
\begin{align}
0&=w'(t)=\frac{\partial F}{\partial x}\bigg|_{(t,\eta(t))} \cdot 1+\frac{\partial F}{\partial y}\bigg|_{(t,\eta(t))}\cdot \eta'(t)
\end{align}
Rearranging this equation, we get
\begin{align}
\eta'(t)&=-\frac{\frac{\partial F}{\partial x}\bigg|_{(t,\eta(t))}}{\frac{\partial F}{\partial y}\bigg|_{(t,\eta(t))}}.
\end{align}
Hopefully with the different notation, it's clear what the different functions are, and how the chain rule is being applied, and where everything is evaluated.
If the $x,y$ are confusing (and I believe they are), you can write the chain rule computation as follows: for each $t\in I$,
\begin{align}
0&=w'(t)=(\partial_1F)_{(t,\eta(t))}\cdot 1+(\partial_2F)_{(t,\eta(t))}\cdot \eta'(t),
\end{align}
and hence
\begin{align}
\eta'(t)&=-\frac{(\partial_1F)_{(t,\eta(t))}}{(\partial_2F)_{(t,\eta(t))}}
\end{align}
It is an abuse of notation to use $y$ to refer to both the coordinate, and also the name of the implicitly defined function, and to use $F$ as both the original function, and the new composed function $w$, but unfortunately, it is standard practice.
Best Answer
Hint:
A level curve of a surface is a curve where $f(x,y)$ is constant. Since we want the level curve that contains $(1,1)$, we plug in this point to get $f(1,1)=4$. So we want to find the line tangent to $$4=3x^2y^2+2x^2-3x+2y^2$$ through the point $(1,1)$. Now, you should use implicit differentiation to find $\frac{dy}{dx}$.
If you are looking to use the partial derivatives instead of the implicit differentiation, for a level curve $F(x,y)=k$, that $$\frac{dy}{dx}=\frac{-F_x}{F_y}.$$
Either way, once you have $\frac{dy}{dx}$, use point-slope form $y-y_0=m(x-x_0)$ to get the equation of the tangent line.