Does the limit as $(x,y) \to (1,2)$ of $3x^3-x^2 y^2$ exist?

The title says it all: does the following limit exist?

$$\lim_{\large(x, y) \to (1, 2)} \; 3x^3 - x^2y^2$$

It approaches $\,-1\,$ with direct substitution, but if you approach the point with the curve $x = y^2$, you get $\,128\,$ which is not $\,-1,\,$ meaning the limit doesn't exist. I got the problem wrong but I am curious to see if it does actually exist.

I'm approaching with $\,x = y^2.\,$ So it ends up being $\;\displaystyle\lim_{y\to 2}\;3y^6-y^6,\,$ which is $128$.

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do you mind showing your computation for the limit along the curve? – Ittay Weiss Feb 12 '13 at 0:59
This function is continuous on $\mathbb{R}^2$, so in particular at $(1,2)$. So... – 1015 Feb 12 '13 at 1:01
What proof do you have of the function being discontinuous at (1,2)? If it is continuous there then the direct substitution is one way to compute the answer. – JB King Feb 12 '13 at 1:03
Our book says to check from all directions and gives an example similar to this saying that the limit approaches the same thing from the x direction and the y direction but then why computing for y^2 it gets a different answer concluding that the limit DNE – user61918 Feb 12 '13 at 1:05
If $y \to 2$ and $x=y^2$ then $x \to 4$... So the second limit you are calculating is $\lim_{(x,y) \to (4,2)}3x^3-x^2y^2$ along some particular curve.... – N. S. Feb 12 '13 at 1:30

$(x, y) \to (1, 2),\;$ but $\;(1, 2) \notin x=y^2$
The function is continuous on $\mathbb{R}^2,\;$ hence at $(1,2)$, so direct substitution is suffices to compute the limit, as you started out doing.
If the curve you are following along is $x=y^2$ then if $x=1$, we get $y=1$ on the curve. Likewise if $y=2$ then $x=\sqrt{2}$. You cannot approach point $(1,2)$ along the curve $x=y^2$