Show that $\boldsymbol{\mathrm{F}}$ is independent of path. 
Consider a vector field $\boldsymbol{\mathrm{F}}(x,y) = \langle 2xy,
 x^2 \rangle$ and three curves that start at $(1, 2)$ and end at
  $(3,2)$. Explain why  $$\int\limits_{C}\boldsymbol{\mathrm{F}}\cdot
 \text{ d}\boldsymbol{\mathrm{r}}$$ has the same value for all three
  curves, and what is this common value?

(There is a graph of three curves, but I'm pretty sure it's not necessary. For your reference, this is Stewart's Calculus, p. 1082, section 16.3 #11.)
My work: notice that 
$$\begin{align}
&\dfrac{\partial}{\partial y}[2xy]=2x \\
&\dfrac{\partial}{\partial x}[x^2] = 2x
\end{align}$$
and thus, $\boldsymbol{\mathrm{F}}$ is conservative.
My understanding is that we need to show that $\boldsymbol{\mathrm{F}}$ is independent of path. Looking at my theorems provided doesn't help. 
And if I do find such a theorem, I'm not sure what to use for $\boldsymbol{\mathrm{r}}$.

As you showed, the vector field is conservative, so it doesn't matter which path you take, the only thing you need are the starting and end point.
First, as $\mathrm{F}$ is conservative, you have to calculate a function $f$ such that $\nabla f=\mathrm{F}$. An easy way to do this is using this formula:
$$\displaystyle f(x,y) = \int_{0}^{x}\mathrm{F}_{1}(t,0)dt + \int_{0}^{y}\mathrm{F}_2(x,t)dt$$
Where $\mathrm{F}_1,\mathrm{F}_2$ are the first and second value of the vector field $\mathrm{F}$. Therefore $$\displaystyle \int_{C}\mathrm{F}\cdot dr = f(x,y)|^{r_1}_{r_0}$$
Where $r_1$ and $r_0$ are your end and starting points, respectively.
