Prove that $\lim_{(x,y)\to(0,0)}\frac{x^2+y^2+5xy}{x-y}$ does not exist $$\lim_{(x,y)\to(0,0)}\frac{x^2+y^2+5xy}{x-y}$$
I've tried approximating using parabolas and even use polar coordinates but it seems to always result in $0$.
The limit seems simple so I must be forgetting something basic. 
Hints? Thank you.

A coordinate transformation can work - but you just have to get a little bit more ingenious. The other answers have shown that the trick is the line $x = y$, so we should choose a transformation which makes analyzing it a bit more palatable. Consider the transformation to a tilted coordinate system in which the $y$-axis lies along this line. There are many ways to do this, one of which is the transformation matrix
$$\mathbf{T} := \begin{bmatrix} 1 && -1 \\ 1 && 1 \end{bmatrix}$$
Geometrically, this matrix is the rotation of the coordinate system around the origin counterclockwise by 45 degrees, thus turning the line $x = y$ counterclockwise just the right amount to align with the $y$-axis (as it lies at a 45-degree angle) multiplied by a scaling by $\sqrt{2}$ to make the entries integer and so easy to work with. Under this transformation, a coordinate vector $(x, y)$ transforms as
$$(x, y) \mapsto (x - y, x + y)$$
That is, define $u := x - y$ and $v := x + y$ as your new coordinates. Then the denominator becomes $u$, and the numerator is solved as follows: note that $v^2 = (x^2 + 2xy + y^2)$ so
$$x^2 + y^2 + 5xy = (x^2 + 2xy + y^2) + 3xy = v^2 + 3xy$$
and now, by recognizing the sum-and-difference-of-the-same-thing pattern, that you can take $u + v = 2x$ and $u - v = -2y$ so $(u + v)(u - v) = -4xy$ and $\frac{3}{4} (u + v)(u - v) = 3xy$. Thus
$$\begin{align}\frac{x^2 + y^2 + 5xy}{x - y} &= \frac{v^2 - \frac{3}{4}(u + v)(u - v)}{u} \\
&= \frac{v^2 - \frac{3}{4}(u^2 - v^2)}{u} \\
&= \frac{v^2 - \frac{3}{4}u^2 + \frac{3}{4}v^2}{u}\\
&= \frac{-\frac{3}{4}u^2 + \frac{7}{4}v^2}{u}\\
&= -\frac{3}{4} u + \frac{7}{4} \frac{v^2}{u} \\
&= \frac{7}{4} \frac{v^2}{u} - \frac{3}{4} u
\end{align}$$
and now when you take the limit - note that the origin is unchanged, so the limit is still to $(0, 0)$ but now in $(u, v)$ coordinates instead of $(x, y)$ coordinates - the second term is zero, but the first has no limit: if $v$ goes to 0 first, then the limit as $u$ goes to 0 is 0, but if $u$ goes to 0 first, the limit becomes $\pm \infty$, and is such for every value of $v$ except 0 for such an approach. That is, the function is unboundedly large in an arbitrarily small circle around the origin, and thus the limit at that origin does not exist.
