Consider $A = {f(x) \in \mathbb{R}(x):f'(0)=0}$. Is $A$ an ideal in $\mathbb{R}(x)$?
My answer: No it is not an ideal. Consider $g(x)\in \mathbb{R}(x)$ s.t. $g(x) = x$. And $f(x)\in A$ s.t. $f(x)= 1$. $g(x)f(x) = x$ and the $\frac{d}{dx} x = 1 \neq 0$. So it is not the case that for any $r \in \mathbb{R}(x)$ and any $x \in A$, $xr \in A$. So A is not an ideal.
Did I interpret the definition of an ideal correctly?
I think it is! An ideal is closed under multiplication by any elements in the ring. That is not closed under multiplication by any elements in the ring, hence not an ideal.