= Solution
The parametrization lands on the curve because $(t^3)^4=(t^4)^3$. Its <pullback> has <image>
$$
\operatorname{im}F^\#=k[t^3,t^4]\subset k[t]=A(\mathbb A^1).
$$
Every <monomial> in this subalgebra has exponent $3a+4b$ for nonnegative integers $a,b$. No such exponent is one, so no <linear combination> of these <monomials> is $t$. Thus $F^\#$ is not <surjective> and \b[$F$ is not an <isomorphism>].
It is nevertheless <bijective> on <points>. Away from the origin, $x\neq0$ and $t=y/x$ gives $t^3=x$ and $t^4=y$ using $x^4=y^3$; the origin comes only from $t=0$. The rational inverse cannot extend regularly at the origin: if $t=P(t^3,t^4)/Q(t^3,t^4)$ with $Q(0)\neq0$, then $tQ(t^3,t^4)$ has a nonzero degree-one term, whereas $P(t^3,t^4)$ has none.
$$
\boxed{F\text{ is bijective, but its inverse is not regular at the cusp}.}
$$
This distinguishes a set-theoretic <bijection> from an <isomorphism> of <varieties> in every <characteristic>.
\Image[/past-exam-of-the-algebraic-geometry-course-of-the-university-of-oxford/2011-cusp-parametrizations.png]
{title=Cusp parametrizations reveal which rational coordinates fail to be regular}
{description=The left curve uses x=t cubed and y=t to the fourth; recovering t fails regularly at its origin. The right curve uses x=t cubed and y=t squared; the open subset used later omits the parameter values zero and one, marked with hollow circles.}
This is a <bijective monomial cusp parametrization with nonregular inverse>.
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