Let $C = {\{(t,t^2,t^3): t \in K\}} = V(y-x^2, z-xy) \subset A^3$ be the affine twisted cubic curve.
Compute the homogenizations $\bar f_1$ and $\bar f_2$ if the polynomials $f_1 = y -x^2$ and $f_2 = z - xy$ and show $\bar C \subsetneq V(\bar f_1, \bar f_2)$. attempt: let $I = <f_1,f_2> = <y-x^2, z-xy>$. Then we dehomegenize with respect to $x_i $ define by $f: K[x_0,...,x_n] \rightarrow K[x,y,z]$. I have no idea how to continue. I now $wy - x^2$ and $wz - xy$ work, but I am not sure how to get them. Also the other part. Could someone please help? thank you.
I don't understand what you're trying to do.
Homogenizing polynomials just mean that you multiply every monomial by enough powers of $w$ such that the polynomial is homogeneous. As you say, the homogenization of $y-x^2$ is $wy-x^2$: the polynomial is of degree $2$, so we multiply all terms by enough $w$'s so that every term has degree $2$. In this case it is enough to multiply only the first term by $w$.
Another example Let $f=x^3+x^2-y$. The homogenization (with respect to $z$) is $f=x^3+x^2z-yz^2$.
Now to the second part: We want to show that $\bar C \subsetneq V(\bar f_1, \bar {f_2})$. This means that there is some point in $ V(\bar f_1, \bar {f_2})$ that is not on $\bar C = \{ (s^3:s^2t:st^2:t^3) \} \subset \mathbb P^3$.
The strategy is to consider what new points we get when compactifying: we have that $\mathbb P^3 \backslash \mathbb A^3 = V(w = 0)$. Putting $w=0$, we find that we must have $x=0$, and $y$ and $z$ can vary freely. Thus all points on the projective line $\{ (0:y:z:0) \}$ lie in $ V(\bar f_1, \bar {f_2})$, but not all points on this line lies on the twisted cubic.