Let $I=(y^2-x^2,y^2+x^2)$. Find $V(I)$ and $\text{dim}_{\mathbb{C}}(\mathbb{C}[x,y]/I)$ To find $V(I)$ we note that the only point $(x,y)\in \mathbb{C}$ satisfying both $y^2-x^2=0$ and $y^2+x^2=0$ is $(0,0)$, so $V(I)=(0,0)$. Now, to find  $\text{dim}_{\mathbb{C}}(\mathbb{C}[x,y]/I)$, we have the following relations:
$$y^2-x^2\equiv 0 \mod I$$
$$y^2+x^2\equiv 0 \mod I$$
From where it follows that $y^2 \equiv 0 \mod I$, and consequently $x^2\equiv 0 \mod I$. From this, I understand that every monomial with degree 3 or more, like $xy^2$ for example, is $0$. The polynomials in $\mathbb{C}[x,y]/I$ then are just a linear combination of $\lbrace 1,x,y,xy \rbrace $, and so $\text{dim}_{\mathbb{C}}(\mathbb{C}[x,y]/I)=4$. Is this correct? I would like to know a more general way of doing this, given that this felt like a very rudimentary approach, and only possible given that the ideal was very simple. If I had any other ideal generated by a couple of polynomials in $\mathbb{C}[x,y]$, how could I find $\mathbb{C}[x,y]/I$? This is a problem from William Furton's Algebraic Curves, and I think the author's intention was for the student to find a most sophisticated approach. 
 A: Edit: As Georges mentions in the comments, I didn't notice that we are finding the dimension of $\mathbb{C}[x, y]/I$ as a vector space over $\mathbb{C}$ as opposed to its Krull dimension as a quotient ring. In that case, I think your reasoning is correct, yes. This agrees with Corollary 4 from Fulton, in that the number of points of $V(I)$, one, is less than $\dim_\mathbb{C} (\mathbb{C}[x, y]/I) = 4$.
I'll leave the old answer about Krull dimension below in case anyone is interested.

Old answer about Krull dimension: We have the result that for any Noetherian ring $A$, if $f \in A$ is neither a zero-divisor nor unit then $\mathrm{ht}((f)) = 1$. It follows that $\mathrm{ht}((f)) = 1 \implies \dim(A/(f)) \leq \dim A -1$.
In your case, we can write 
$$  \mathbb{C}[x, y]/(y^2-x^2, y^2+x^2) \cong [ \mathbb{C}[x, y]/(y^2 - x^2) ] / [(y^2 - x^2, y^2 + x^2)/(y^2 - x^2)] . $$
To do this I've used the fact at the bottom of this question. Then since $(y^2 - x^2, y^2 + x^2)/(y^2 - x^2) = (\overline{y^2 + x^2})$ (generated by the image of $y^2 + x^2$ in $\mathbb{C}[x, y]/(y^2-x^2)$) is neither a unit nor zero-divisor in $\mathbb{C}[x, y]/(y^2 - x^2)$ (though I have not explicitly checked this; please let me know if there are any problems with this!) and $y^2 - x^2$ is neither a unit nor zero-divisor in $\mathbb{C}[x, y]$, we get 
\begin{align}  \dim (\mathbb{C}[x, y]/I) &\leq  \dim(\mathbb{C}[x, y]/(y^2 - x^2)) - 1 \\ &\leq \dim \mathbb{C}[x, y] - 1 - 1 \\ &= 2 - 1 - 1 \\ &=0 \end{align}
i.e. $\dim (\mathbb{C}[x, y]/I) = 0$.
As a reality check, we should have that the (Krull) dimension of the coordinate ring $k[X]$ of an affine variety $X$ over a field $k$ and the geometric dimension of the variety are equal. You've pointed out that your variety $V(I)$ is just the point $(0, 0)$ (which has dimension $0$) and we have just seen that $\mathbb{C}[V(I)] = \mathbb{C}[x, y]/I$ (the coordinate ring of your variety $V(I)$) has Krull dimension $0$.
