The trick is to replace $y$ by $x^2$ and $z$ by $x^3$ in the quotient ring $k[x,y,z]/\langle y-x^2, z-x^3\rangle$ so that the quotient becomes:
$k[x,y,z]/\langle y-x^2, z-x^3\rangle =k[x,x^2,x^3]=k[x]$.
Hence, since the quotient is a domain, your wish has come true : the ideal $\langle y-x^2, z-x^3\rangle$ is prime.
If you have a quotient of a polynomial ring $k[x_0,x_1,..., x_n]$ of the special form
$$A=k[x_0,x_1,\dots, x_n]/\langle x_0-g(x_1,\dots,x_n),f_1(x_0,\dots, x_n),\dots,f_i(x_0,\dots, x_n),\dots\rangle$$ you can apply the same trick of replacing every occurrence of $x_0$ by $g(x_1,\dots,x_n)$ thus obtaining
$$A=k[x_1,\dots, x_n]/\langle f_1(g(x_1,\dots,x_n),\dots, x_n),\dots,f_i(g(x_1,\dots,x_n),\dots, x_n),\dots\rangle.$$ In the example you submitted I implicitly used that twice (but only wrote the end result), exploiting the rule for successive quotients
$$k[x,y,z]/\langle f,g\rangle=(k[x,y,z]/\langle f\rangle)/\langle \bar g\rangle.$$ Edit: a geometric interpretation
We have an embedding $\mathbb A^1 \to \mathbb A^3: t\mapsto (t,t^2,t^3)$ with image an irreducible afine variety $ V$ known as the affine twisted cubic curve.
The calculation above says that this variety is the ideal-theoretic (or scheme-theoretic) intersection of the affine quadrics $y=x^2$ and $z=x^3$.
It is interesting to notice that the closure $\bar V$ of $V$ in $\mathbb P^3$ is no longer a scheme-theoretic complete intersection even though it is set-theoretically the intersection of a quadric and a cubic surface.