What are the irreducible components of $V(xy-z^3,xz-y^3)$ in $\mathbb{A}^3_K$? 
What are the irreducible components of the algebraic set $V(xy-z^3,xz-y^3)$ in $\mathbb{A}^3_K$? (Here I'm letting $K$ be an algebraically closed field.)

Normally, what I do is take the equations determining an algebraic set $V(I)$, and usually one of them factors so that $V(I)$ decomposes as $V(J_1)\cup V(J_2)\cup\cdots$ or something. After breaking things down enough, I can eventually find that $K[x,y,z]/J_i$ is an intergral domain, so $J_i$ is prime, and $V(J_i)$ is irreducible. 
However, I don't see any obvious way to break $V(xy-z^3,xz-y^3)$ down as a union of smaller sets since neither of the polynomials factor. I think  $K[x,y,z]/(xy-z^3,xz-y^3)\cong K[x]\oplus \langle y,z\rangle$ because any occurrence of $xy$ or $xz$ be be replaced with a power of $z^3$ or $y^3$, but I'm not sure if that helps.
 A: Let $(x,y,z) \in V(xy-z^3,xz-y^3)$. Two cases:
$1$. $x=0$. Then $y=z=0$.
$2$. $x\ne 0$. Then $y^4 = z^4$, so $z = \omega \cdot y$ where $\omega^4 = 1$. If $y=0$ then $z=0$. If $y \ne 0$ then $z \ne 0$ and we have $(x,y,z) = (\omega^3 y^2, y, \omega y)$.
Therefore we have one component, the line $y=z=0$, and for each $\omega$, a fourth root of $1$, we have a component isomorphic to $\mathbb{A}^1$: $\{ (\omega^3 t^2, t, \omega t )\ | \ t \in k \}$.
A: If you take the two relations $xy-z^3$ and $xz-y^3$ and add them together, you can rewrite their sum as $x(y+z)-(z^3 + y^3) = z^2(y+z)(x-z^2+y)$, after using the fact that $xy = z^3$. As a consequence, we can write $V(xy-z^3, xz-y^3)$ as
$$
V(xy-z^3,xz-y^3,y+z) \cup V(xy-z^3, xz-y^3, z^2) \cup V(xy-z^3, xz-y^3, x-z^2+y)
$$
Now each of these 3 algebraic sets should be easier to decompose - I'll work out the left one here. 
\begin{align*}
V(xy-z^3, xz-y^3, y+z)
&= V(xy-(-y)^3, x(-y)-y^3, y+z)\\
&=V(xy+y^3, y+z)\\
&= V(x+y^2, y+z) \cup V(y,y+z) = V(x+y^2,y+z) \cup V(y,z).
\end{align*}
However, both $K[x,y,z]/(x+y^2,y+z) \simeq K[y]$ and $K[x,y,z] / (y,z) \simeq K[x]$ are integral domains, hence $V(x+y^2,y+z)$ and $V(y,z)$ are irreducible. To get the other irreducible components, we would use a similar procedure to break up $V(xy-z^3,xz-y^4,z^2)$ and $V(xy-z^3,xz-y^3,x-z^2+y)$ into irreducibles.
A: This is to complemenent msteve's excellent answer.
Following his procedure would yield potentially $3 \times 2=6$ irreducible components. However, $(y,z)$ shows up in all of them, so we get only $4$, and in fact these are all.
This can be computed in Macaulay2 as follows:
i1 : R = QQ[x,y,z]
o1 = R
o1 : PolynomialRing
i2 : I = ideal(x*y-z^3, x*z-y^3     
               3           3
o2 = ideal (- z  + x*y, - y  + x*z)
o2 : Ideal of R

i3 : decompose I

              2    2                                         2
o3 = {ideal (y  + z , y*z + x), ideal (z, y), ideal (y + z, z  + x), ideal (- y
 ---------------------------------------------------------------------------
         2
 + z, - z  + x)}

o3 : List

(I've removed some of the output for readability). 
We see that this answer agrees with msteve's hand computations.
EDIT  I should be more critical of what my computer tells me! Of course, $\mathbb Q$ is not algebraically closed, so we don't see all irreducible components. A quick glance at $(y^2+z^2,yz+x)$ reveals that this is not irreducible ($y^2+z^2=(y+zi)(y-zi)$), so we do in fact get $5$ components! 
Running the computation again, but this time over $\mathbb Z/101$, gives a different answer:
i8 : decompose I

                             2                                    2
o8 = {ideal (- 10y + z, - 10z  + x), ideal (z, y), ideal (y + z, z  + x), ideal
     ---------------------------------------------------------------------------
                  2                          2
     (- y + z, - z  + x), ideal (10y + z, 10z  + x)}

o8 : List

