Let's look at the cuspidal cubic curve $C: x^3 = y^2$. ~~I claim $H^1_{dR}(C)$ is two-dimensional.~~

**EDIT**: Actually $H^1_{dR}(C) = 0$, so that this curve does **not** give an example of an affine singular variety whose algebraic de Rham cohomology differs from its singular cohomology. This is corroborated by a remark in the beginning of section 3.2 of Huber and Müller-Stach's Periods and Nori Motives (just control-F for "cusp"). Note that the remark is not in their first version, but it is in their sixth which can be found on Prof. Huber's website.

Thanks are due to Julian Rosen for both pointing out a mistake in the derivation defined in the original version as well as realizing the important and subtle point that $\Omega^2_C \neq 0$, being supported at the origin. My sincere apologies for posting prematurely to the (three or more) people who read this answer in the meantime. The derivation as defined originally was not a derivation, but can be replaced by another which serves the same purpose as the original:

$$ \frac{\mathbb C[x,y]}{(x^3-y^2)} \to \frac{\mathbb C[x]}{(x^3)} : x \mapsto x, y\mapsto 1 $$

This derivation still shows that $\eta$ and $x\eta$ are non-zero. The more important point is that the presence of $\Omega^2_C$ means that the forms $x\,dy$ and $x^2\,dy$ are not actually closed, so these forms do not contribute to cohomology as was incorrectly assumed before, because $\Omega^2_C$ is in fact two-dimensional and spanned by $d(x\,dy) = dx\wedge dy$ and $d(x^2\,dy) = 2x\,dx\wedge dy$. These 2-forms are zero everywhere but at the singularity. My assumption that they were closed was the mistake which led me to conclude the cohomology was two-dimensional. The computations and formulas in the post show that the other 1-forms are exact so that $H^1_{dR}(C) = 0$. At any rate this mistake is an instructive one I think, so I'll preserve the post as originally written. Again, the two mistakes of this post are in the definition of the derivation and the incorrect assumption that $x\,dy$ and $x^2\,dy$ are closed forms.

As Georges notes, the singular cohomology of its closed points is trivial.

Regardless of whether it's direct (it's not), we definitely have the sum
$$\Omega^1_C = \mathbb C[x]\,dx + \mathbb C[x]y\,dx + \mathbb C[x]\,dy + \mathbb C[x]y\,dy $$

and then because $3x^2\,dx = 2y\,dy$, the last summand is actually unnecessary. The first summand is obviously composed of exact forms, so the only forms in question are things like $x^ny\,dx$ and $x^n\,dy$ where $n\geq 0$. On the other hand,

$$ d\left( \frac{2x^{n+1}y}{2n+5} \right) = x^ny\,dx \quad \text{ for } n > 1 $$
$$d\left( \frac{3 x^ny}{2n+3} \right) = x^n \,dy \quad \text{ for } n > 2$$

For the lower $n$, one needs to divide by $x$'s and $y$'s which is only valid provided $\Omega^1_C$ is torsion-free at the origin (and later we'll see it isn't), so these need to be handled with care. This leaves $y\,dx$, $xy\,dx$, $x\,dy$, and $x^2\,dy$. As $d(xy) = x\,dy + y\,dx$, any $\mathbb C$-linear relation between $y\,dx$ and $x\,dy$ (except for multiples of $y\,dx = -x\,dy$) would show that they are both exact. Similarly, $d(x^2y) = 2xy\,dx + x^2\,dy$ so any $\mathbb C$-linear relation between $xy\,dx$ and $x^2\,dy$ (except for multiples of $2xy\,dx = -x^2\,dy$) would show that they are both exact.

The obvious place to look is the relation $3x^2\,dx = 2y\,dy$, which after multiplying with $y$ gives $$ x^2(3y\,dx - 2x\,dy) = 0 $$ as well as $$ x(3xy\,dx - 2x^2\,dy) = 0 $$ Hence the form $\eta = 3y\,dx - 2x\,dy$ is very relevant: if $\eta = 0$ then we get both linear relations from $\eta = x\eta = 0$ and then we would have shown that the first cohomology group is trivial.

More subtly, it turns out $\eta$ is not zero. What the first formula does show is that it is zero in any localisation of $\Omega^1_C$ away from $(0,0)$. This funny form also maps to zero in the cotangent space at zero, since $\eta \in (x,y)\Omega^1_C$. This implies that any *first-order* derivation cannot determine whether $\eta \neq 0$.

~~To see that $\eta \neq 0$ we will use the ~~*second-order* derivation (the correct derivation is above in the comment) $$ \frac{\mathbb C[x,y]}{(x^3-y^2)} \to \frac{\mathbb C[x,y]}{(x^2,y^2)} : x,y\mapsto 1, \, x^2,y^2 \mapsto 0$$ The universal property of $\Omega^1_C$ then promises that there's a unique map of $\frac{\mathbb C[x,y]}{(x^3-y^2)}$-modules $$ \Omega^1_C \xrightarrow{f} \frac{\mathbb C[x,y]}{(x^2,y^2)} : dx,dy\mapsto 1 $$ Since $f(\eta) = 3y-2x$ and $f(x\eta) = 3xy$ which are both non-zero, we've shown $\eta,x\eta \neq 0$. One way of thinking about $x\eta$ is that it is a Dirac delta supported at $(0,0)$, while $\eta$ is a (weak) derivative of a Dirac delta supported at $(0,0)$.

Now working modulo exact forms, we've shown that $[ydx] = -[xdy]$ and $[x^2\,dy]=-2[xy\,dx]$ in $H^1_{dR}(C)$. I claim these are non-zero and independent.

Following up on David's comment we weight $\frac{\mathbb C[x,y]}{(x^3-y^2)}$ so that $\text{wt }x = 2$ and $\text{wt }y = 3$ and then weight $\Omega_{A/k}$ accordingly in order that $d$ preserves weight (The motivation for this is that we are measuring in units of $t$ under the rationalization map $t \mapsto (t^2,t^3)$).

Because $\eta$ (resp., $x\eta$) is not zero, the weight 5 piece (resp., weight 7) of $\Omega_{C}$ is two-dimensional, spanned by $y\,dx$ and $x\,dy$ (resp., $xy\,dx$ and $x^2\,dy$). On the other hand, the weight 5 piece (resp., weight 7) of $\frac{\mathbb C[x,y]}{(x^3-y^2)}$ is only one-dimensional, spanned by $xy$ (resp., $x^2y$).

~~We conclude that $H^1_{dR}(C)$ is two-dimensional, supported in weights 5 and 7 and spanned by $[xdy]$ and $[x^2\,dy]$, respectively.~~ See edit comment above.