The question has three parts. I think i figured out the first two, but i have no idea on how to do the last.
a) Let $Y$ be the plane curve $y=x^2$. Show that $\frac{k[x,y]}{I(Y)}$ is isomorphic to a polynomial ring in one variable over $k$.
Ok, so in $\frac{k[x,y]}{(f)}$ we can identify $y$ with $x^2$, making the ring isomorphic to $k[x]$.
b) Let $Z$ be the plane curve $xy = 1$. Show that $\frac{k[x,y]}{I(Z)}$ is NOT isomorphic to a polynomial ring in one variable over $k$.
$(xy-1)$ is maximal over $k[x,y]$, so $\frac{k[x,y]}{I(Z)}$ is a field. A polynomial in one variable over a field doesn't have an inverse for the variable, so it cannot be a field.
c) Let $f$ be any irreducible polynomial in $k[x,y]$,and let $W$ be the conic defined by $f$. Show that $\frac{k[x,y]}{I(W)}$ is isomorphic to $\frac{k[x,y]}{I(Y)}$ or $\frac{k[x,y]}{I(Z)}$.
I don't have any idea on how to do this.
This is the very first exercise from Algebraic Geometry by Hartshorne. There are some solutions online.
a) as the author writes, one can consider the morphism $\phi: A(Y) \rightarrow k[x],\ \phi([f(x,y)])=f(x,x^2)$ and prove that it is linear, surjective and injective (as $\mathtt{Ker} \phi=\{[0]\}$).
b) I found three ideas one can use to show this:
In $k[x]$ the only invertible elements are the constants, while in $A(Z)$ both $x$ and $y$ are invertible.
Elements of the type $a_{rs}x^ry^s$, when considered $\mod{xy-1}$ are equivalent to $a_{rs}x^{r-s}$ if $r\geq s$ or to to $a_{rs}y^{s-r}$ if $r<s$
As in a), you can show that $K[x,y]/(xy-1) \cong K[x,x^{-1}] \ncong K[x]$
c) The author missed "quadratic" in the problem. Another author suggested the right path in a comment: write $f(x,y)=ax^2+by^2+cxy+dx+ey+g$ into one of two forms above.