You should really make this question more clear. What do you mean by a "partial resolution"? For instance, is it normal, or at least $S_2$? Is it $\mathbb Q$-factorial? Or is at least the canonical divisor $\mathbb Q$-Cartier? You need some assumption for the question to make sense.

First let's take $n=2$. Blowing up the singular point gives you the crepant resolution $\pi:Y\to X$. (By the way if a surface has a crepant resolution it must be the minimal resolution and hence it is unique). So, let's blow up a point on the $\pi$-exceptional curve $E\subset Y$: $\sigma: Z\to Y$. Now $Z$ is smooth and $F=\sigma^{_1}_*E$ has self-intersection $F^2=-3$ and $F\simeq \mathbb P^1$, so it can be contracted to a normal surface $Z\to W$. By construction we get a proper birational morphism $W\to X$. This is a partial resolution, but it is not crepant.

So, perhaps you mean a partial resolution that's partial to the crepant resolution. I mean that it is an intermediate partial resolution. In the case $n=2$ there is not much room for that, but in general that kind of partial resolutions are also crepant by the following Lemma (but let's do a definition first).

Definition Let $\alpha:X\to Y$ be a proper birational morphism between normal varieties. Assume that $K_Y$ is $\mathbb Q$-Cartier, then $\alpha$ will be called crepant if $K_X\sim \alpha^* K_Y$.

Lemma Let $\alpha:X\to Y$ and $\beta:Y\to Z$ be two proper birational morphisms and assume that $X,Y,Z$ are normal algebraic surfaces such that $K_Y$ and $K_Z$ are $\mathbb Q$-Cartier. then if $\beta\circ\alpha$ is crepant, then so are $\alpha$ and $\beta$.

Proof: Write $$K_X=\alpha^*K_Y+\sum a_iE_i$$ where $E_i$ are $\alpha$-exceptional irreducible divisors and $$K_Y=\beta^*K_Z+\sum b_jF_j$$ where $F_j$ are $\beta$-exceptional irreducible divisors. It follows that then $$K_X=\alpha^*\beta^*K_Z+\sum a_iE_i+\sum b_j\alpha^* F_j.$$ If $\beta\circ\alpha$ is crepant, then $$\sum a_iE_i+\sum b_j\alpha^* F_j\sim 0.$$ However, for each $j$, we have that $\alpha^*F_j= \widetilde F_j +\sum a_{ij}E_i$ where $\widetilde F_j$ is the strict transform of $F_j$ on $X$, so we get that for some other set of coefficients we have $$\sum c_iE_i+\sum b_j\widetilde F_j\sim 0.$$ The $E_i$ and $\widetilde F_j$ together are all $\beta\circ\alpha$ exceptional and hence linearly independent. Therefore $b_j=0$ for all $j$ and then $$\sum a_iE_i\sim 0$$ with the original $a_i$'s. The same reasoning says that then the $a_i=0$ which proves the statement.$\square$

Remark
It seems to me that the same proof works in arbitrary dimension, so one does not need to limit to surfaces for the Lemma.

You should really make this question more clear. What do you mean by a "partial resolution"? For instance, is it normal, or at least $S_2$? Is it $\mathbb Q$-factorial? Or is at least the canonical divisor $\mathbb Q$-Cartier? You need some assumption for the question to make sense.

Consider the crepant resolution $\pi:Y\to X$. (By the way if a surface has a crepant resolution it must be the minimal resolution and hence it is unique). Next, let's blow up a point on a $\pi$-exceptional curve $E\subset Y$ (say one that's not contained in any other exceptional curves) $\sigma: Z\to Y$. Now $Z$ is smooth and $F=\sigma^{-1}_*E$ has self-intersection $F^2=-3$ and $F\simeq \mathbb P^1$, so it can be contracted to a normal surface $Z\to W$. By construction we get a proper birational morphism $W\to X$. This is a partial resolution, but it is not crepant.

So, perhaps you mean a partial resolution that's partial to the crepant resolution. I mean that it is an intermediate partial resolution. That kind of partial resolutions are also crepant by the following Lemma (but let's do a definition first).

Definition Let $\alpha:X\to Y$ be a proper birational morphism between normal varieties. Assume that $K_Y$ is $\mathbb Q$-Cartier, then $\alpha$ will be called crepant if $K_X\sim \alpha^* K_Y$.

Lemma Let $\alpha:X\to Y$ and $\beta:Y\to Z$ be two proper birational morphisms and assume that $X,Y,Z$ are normal algebraic surfaces such that $K_Y$ and $K_Z$ are $\mathbb Q$-Cartier. then if $\beta\circ\alpha$ is crepant, then so are $\alpha$ and $\beta$.

Proof: Write $$K_X=\alpha^*K_Y+\sum a_iE_i$$ where $E_i$ are $\alpha$-exceptional irreducible divisors and $$K_Y=\beta^*K_Z+\sum b_jF_j$$ where $F_j$ are $\beta$-exceptional irreducible divisors. It follows that then $$K_X=\alpha^*\beta^*K_Z+\sum a_iE_i+\sum b_j\alpha^* F_j.$$ If $\beta\circ\alpha$ is crepant, then $$\sum a_iE_i+\sum b_j\alpha^* F_j\sim 0.$$ However, for each $j$, we have that $\alpha^*F_j= \widetilde F_j +\sum a_{ij}E_i$ where $\widetilde F_j$ is the strict transform of $F_j$ on $X$, so we get that for some other set of coefficients we have $$\sum c_iE_i+\sum b_j\widetilde F_j\sim 0.$$ The $E_i$ and $\widetilde F_j$ together are all $\beta\circ\alpha$ exceptional and hence linearly independent. Therefore $b_j=0$ for all $j$ and then $$\sum a_iE_i\sim 0$$ with the original $a_i$'s. The same reasoning says that then the $a_i=0$ which proves the statement.$\square$

Remark
It seems to me that the same proof works in arbitrary dimension, so one does not need to limit to surfaces for the Lemma.

You should really make this question more clear. What do you mean by a "partial resolution"? For instance, is it normal, or at least $S_2$? Is it $\mathbb Q$-factorial? Or is at least the canonical divisor $\mathbb Q$-Cartier? You need some assumption for the question to make sense.

Actually, I don't know how this $X=\mathbb C^2/\mathbb Z_n$ has a crepant resolution unless $n=2$... Perhaps I am misunderstanding the situation, but I think that the minimal resolution has a single exceptional curve with discrepancy $-1+\frac 2n$. Or perhaps the $\mathbb Z_n$ action is not what I think it is? If so, you should really clarify what's happening.

But let's take $n=2$.

Depending on what you mean by a partial resolution, this might not be true. Here is an example. If I am not mistaken, then blowing up the singular point gives you the crepant resolution $\pi:Y\to X$. (By the way if a surface has a crepant resolution it must be the minimal resolution and hence it is unique). So, let's blow up a point on the $\pi$-exceptional curve $E\subset Y$: $\sigma: Z\to Y$. Now $Z$ is smooth and $F=\sigma^{_1}_*E$ has self-intersection $F^2=-3$ and $F\simeq \mathbb P^1$, so it can be contracted to a normal surface $Z\to W$. By construction we get a proper birational morphism $W\to X$. This is a partial resolution, but it is not crepant.

So, perhaps you mean a partial resolution that's partial to the crepant resolution. I mean that it is an intermediate partial resolution. In this particular case there is not much room for that, but in general that kind of partial resolutions are also crepant by the following Lemma (but let's do a definition first).

Definition Let $\alpha:X\to Y$ be a proper birational morphism between normal varieties. Assume that $K_Y$ is $\mathbb Q$-Cartier, then $\alpha$ will be called crepant if $K_X\sim \alpha^* K_Y$.

Lemma Let $\alpha:X\to Y$ and $\beta:Y\to Z$ be two proper birational morphisms and assume that $X,Y,Z$ are normal algebraic surfaces such that $K_Y$ and $K_Z$ are $\mathbb Q$-Cartier. then if $\beta\circ\alpha$ is crepant, then so are $\alpha$ and $\beta$.

Proof: Write $$K_X=\alpha^*K_Y+\sum a_iE_i$$ where $E_i$ are $\alpha$-exceptional irreducible divisors and $$K_Y=\beta^*K_Z+\sum b_jF_j$$ where $F_j$ are $\beta$-exceptional irreducible divisors. It follows that then $$K_X=\alpha^*\beta^*K_Z+\sum a_iE_i+\sum b_j\alpha^* F_j.$$ If $\beta\circ\alpha$ is crepant, then $$\sum a_iE_i+\sum b_j\alpha^* F_j\sim 0.$$ However, for each $j$, we have that $\alpha^*F_j= \widetilde F_j +\sum a_{ij}E_i$ where $\widetilde F_j$ is the strict transform of $F_j$ on $X$, so we get that for some other set of coefficients we have $$\sum c_iE_i+\sum b_j\widetilde F_j\sim 0.$$ The $E_i$ and $\widetilde F_j$ together are all $\beta\circ\alpha$ exceptional and hence linearly independent. Therefore $b_j=0$ for all $j$ and then $$\sum a_iE_i\sim 0$$ with the original $a_i$'s. The same reasoning says that then the $a_i=0$ which proves the statement.$\square$

Remark
It seems to me that the same proof works in arbitrary dimension, so one does not need to limit to surfaces for the Lemma.

You should really make this question more clear. What do you mean by a "partial resolution"? For instance, is it normal, or at least $S_2$? Is it $\mathbb Q$-factorial? Or is at least the canonical divisor $\mathbb Q$-Cartier? You need some assumption for the question to make sense.

First let's take $n=2$. Blowing up the singular point gives you the crepant resolution $\pi:Y\to X$. (By the way if a surface has a crepant resolution it must be the minimal resolution and hence it is unique). So, let's blow up a point on the $\pi$-exceptional curve $E\subset Y$: $\sigma: Z\to Y$. Now $Z$ is smooth and $F=\sigma^{_1}_*E$ has self-intersection $F^2=-3$ and $F\simeq \mathbb P^1$, so it can be contracted to a normal surface $Z\to W$. By construction we get a proper birational morphism $W\to X$. This is a partial resolution, but it is not crepant.

So, perhaps you mean a partial resolution that's partial to the crepant resolution. I mean that it is an intermediate partial resolution. In the case $n=2$ there is not much room for that, but in general that kind of partial resolutions are also crepant by the following Lemma (but let's do a definition first).

Definition Let $\alpha:X\to Y$ be a proper birational morphism between normal varieties. Assume that $K_Y$ is $\mathbb Q$-Cartier, then $\alpha$ will be called crepant if $K_X\sim \alpha^* K_Y$.

Lemma Let $\alpha:X\to Y$ and $\beta:Y\to Z$ be two proper birational morphisms and assume that $X,Y,Z$ are normal algebraic surfaces such that $K_Y$ and $K_Z$ are $\mathbb Q$-Cartier. then if $\beta\circ\alpha$ is crepant, then so are $\alpha$ and $\beta$.

Proof: Write $$K_X=\alpha^*K_Y+\sum a_iE_i$$ where $E_i$ are $\alpha$-exceptional irreducible divisors and $$K_Y=\beta^*K_Z+\sum b_jF_j$$ where $F_j$ are $\beta$-exceptional irreducible divisors. It follows that then $$K_X=\alpha^*\beta^*K_Z+\sum a_iE_i+\sum b_j\alpha^* F_j.$$ If $\beta\circ\alpha$ is crepant, then $$\sum a_iE_i+\sum b_j\alpha^* F_j\sim 0.$$ However, for each $j$, we have that $\alpha^*F_j= \widetilde F_j +\sum a_{ij}E_i$ where $\widetilde F_j$ is the strict transform of $F_j$ on $X$, so we get that for some other set of coefficients we have $$\sum c_iE_i+\sum b_j\widetilde F_j\sim 0.$$ The $E_i$ and $\widetilde F_j$ together are all $\beta\circ\alpha$ exceptional and hence linearly independent. Therefore $b_j=0$ for all $j$ and then $$\sum a_iE_i\sim 0$$ with the original $a_i$'s. The same reasoning says that then the $a_i=0$ which proves the statement.$\square$

Remark
It seems to me that the same proof works in arbitrary dimension, so one does not need to limit to surfaces for the Lemma.