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The curves are clearly elliptical. To be more precise, notice that $$ f(x,y) = (x,y)A\binom xy \quad\text{where}\quad A = \begin{bmatrix}3 & 2\\ 2& 3\end{bmatrix} $$ Infinite many such $A$ exist, the one above is the one which is symmetric and hence diagonalizable. Its eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = 5$, with respective eigenversors $v_1 = \frac{1}{\sqrt 2}(1,-1)$ and $v_2 = \frac{1}{\sqrt 2}(1,1)$.
This means that \begin{align} f(x,y) = & (x,y) \begin{bmatrix}3 & 2\\ 2& 3\end{bmatrix} \binom xy \\ = & (x,y) \frac{1}{\sqrt 2} \begin{bmatrix}1 & 1\\ -1& 1\end{bmatrix} \begin{bmatrix}1 & 0\\ 0& 5\end{bmatrix} \sqrt 2 {\begin{bmatrix}1 & 1\\ -1& 1\end{bmatrix}}^{-1} \binom xy \\ = & (x,y) \begin{bmatrix}\frac{1}{\sqrt 2} & \frac{1}{\sqrt 2}\\ -\frac{1}{\sqrt 2}& \frac{1}{\sqrt 2}\end{bmatrix} \begin{bmatrix}1 & 0\\ 0& 5\end{bmatrix} \begin{bmatrix}\frac{1}{\sqrt 2} & -\frac{1}{\sqrt 2}\\ \frac{1}{\sqrt 2}& \frac{1}{\sqrt 2}\end{bmatrix} \binom xy \end{align} Letting $$\tag{1} \begin{cases} \hat x & = \displaystyle\frac{x-y}{\sqrt 2} \\ \hat y & = \displaystyle\frac{x+y}{\sqrt 2} \\ \end{cases} $$ you have $$ f(\hat x,\hat y) = \hat x^2 + 5\hat y^2 $$ therefore the level curves of $f$ in $(\hat x,\hat y)$ coordinates are solutions to $$ \hat x^2 + 5\hat y^2 = C $$ For $C<0$ you have no solutions, for $C=0$ you have only the origin, while for $C>0$ $$ \frac{\hat x^2}{(\sqrt C)^2} + \frac{\hat y^2}{(\sqrt{C/5})^2} = 1 $$ which are ellipses centered in the origin and with semiaxes $\sqrt C$ and $\sqrt{C/5}$ parallel to $\hat x$ and $\hat y$ respectively. Basically they are all concentrical ellipses with ratio horizonal/vertical axis equal to $1/\sqrt 5$.
Notice that the change of variable $(1)$ is simply a rotation of $\frac{\pi}{4}$, so that in the original coordinates $(x,y)$ your curves are such ellipses rotated by $-\frac{\pi}{4}$.
On substituting $u=x+y$ and $v=x-y$, you get, $$x=(u+v)/2$$ and $$y=(u-v)/2$$
Thus,
$$3x^2 + 4xy + 3y^2 = 3(u+v)^2/4 + (u+v)(u-v) + 3(u-v)^2/4$$
On simplification, it reduces to:
$$f(x,y)=g(u,v)=3(u^2+v^2)/2 + u^2-v^2 =( 5u^2 + v^2)/2$$
which is the equation of an ellipse in the $u-v$ plane.
The transformation is like rotating the $x-y$ axes and scaling it. So the cross section's shape wouldn't change.