Let $f: X \to Y$ be a finite, surjective morphism of smooth, quasi-projective varieties over a field $k$ of characteristic zero. Let $p\in X$.
If $\dim X>0$, then does there necessarily exist a smooth curve $C$ on $X$ such that $f(p)$ is a smooth point of the closure of $f(C)$ in $Y$ ?
That is not true. Let $k$ be an algebraically closed field of characteristic prime to $34$. Let the domain of the morphism be a Zariski open neighborhood of the origin in $\mathbb{A}^2_k$ with coordinates $(s,t)$. Let the target of the morphism be a Zariski open neighborhood of the origin in $\mathbb{A}^2$ with coordinates $(u,v)$. Consider the following quasi-finite morphism, $$f:\mathbb{A}^2\to \mathbb{A}^2, \ \ f^*u = s^3 + t^5, \ f^*v = t^7 + s^{11}.$$
Finiteness on appropriate neighborhoods of the origin. By Bezout's Theorem, the generic degree of this morphism is $5\cdot 11 = 55$. The fiber over the origin is a complete intersection zero-cycle in the affine plane. The length of the component supported at the origin is $3\times 7 = 21$. Since $f^*u$ equals zero on this zero-cycle, the residual zero-cycle has coordinates $(s,t) = (a^5,-a^3)$ for some $a\in k^\times$. Plugging into $f^*v$, also $a^{34}-1 = 0$. This has $34$ distinct solutions, and thus the residual has length $34$. Altogether, the zero-cycle has length $55$. Thus, this quasi-finite, flat morphism is finite when restricted to an appropriate Zariski open neighborhood of the origin in the target and inverse image Zariski open neighborhood of the origin in the domain.
Reduction to power series computations. Consider the origin, $p$, in the domain. Every smooth arc of a curve at $C$ is either (Case I) the graph of a function $t=q(s)$ for $q\in sk[[s]]$, or (Case II) the graph of a function $s=r(t)$ for $r\in tk[[t]].$
Case I. The case $q(s)=0$ gives an arc $f(s) = (s^3,s^{11})$, which is a hypercusp. Thus, $q(s)$ is not identically zero. Therefore $q(s)$ has a power series expansion, $$q(s) = cs^m(1+sw(s)), \ \ c\neq 0, \ m\geq 1, \ w(s)\in k[[s]].$$ This gives, $$f^*u(s,q(s)) = s^3(1+c^5s^{5m-3}(1+sw(s))^5), \ \ f^*v(s,q(s)) = c^7s^{7m}(1+sw(s))^7 + s^{11}.$$ If this is the composition of a ramified morphism of smooth curves, $$\phi:C\to D,$$ followed by an immersion, $$\iota:D\hookrightarrow \mathbb{A}^2,$$ then $\iota^*u$ must be a local coordinate on $D$. Thus, $\iota^*v$ must be a power series in $\iota^*u$, say $$\iota^*v = b(\iota^*u)^n + ... , \ \ b\neq 0.$$ Pulling back this identity by $\phi$ gives the identity, $$c^7s^{7m}(1+sw(s))^7 + s^{11} = bs^{3n}(1+c^5s^{5m-3}(1+sw(s))^5)^n + ...$$ If $m$ equals $1$, the lowest power of $s$ with nonzero coefficient on the left-hand side is $7$, and this is not a multiple of $3$. If $m\geq 2$, the lowest power is $11$, and this is also not a multiple of $3$. Thus, Case I leads to a contradiction.
Case II. The case $r(t)=0$ gives an arc $f(t) = (t^5,t^{7})$, which is a hypercusp. Thus, $r(t)$ is not identically zero. Therefore $r(t)$ has a power series expansion, $$r(t) = ct^m(1+tw(t)), \ \ w(t)\in k[[s]].$$ If $m$ equals $1$, then $r(t)$ is a formal isomorphism, and we can find an inverse function $t=q(s)$. By Case I, this leads to a contradiction. Therefore, $m$ is at least $2$.
This gives, $$f^*u(r(t),t) = t^5(1+c^3s^{3m-5}(1+tw(t))^3), \ \ f^*v(r(t),t) = t^7 + c^{11}t^{11m}(1+tw(t))^{11}.$$ If this is the composition of a ramified morphism of smooth curves, $$\phi:C\to D,$$ followed by an immersion, $$\iota:D\hookrightarrow \mathbb{A}^2,$$ then $\iota^*u$ must be a local coordinate on $D$. Thus, $\iota^*v$ must be a power series in $\iota^*u$, say $$\iota^*v = b(\iota^*u)^n + ... , \ \ b\neq 0.$$ Pulling back this identity by $\phi$ gives the identity, $$t^7 + c^{11}t^{11m}(1+tw(t))^{11} = bt^{5n}(1+c^3t^{3m-5}(1+tw(t))^3)^n + ...$$ The lowest power of $t$ on the left-hand side is $7$. Since $7$ is not a multiple of $5$, this is a contradiction in Case II.
Therefore, there is no smooth arc of a curve $C$ at $p$ such that the restriction of $f$ to $C$ factors as a ramified morphism of smooth curves followed by an immersion..
