Proving singularity of a point in a projective variety. Let $n\in\mathbb{Z}_{>1}$ and $k$ be an algebraically closed field. Let $X\subset\mathbb{P}^{2}_{k}$ be the curve given by $x_{1}^{n} = x_{2}x_{0}^{n-1} - x_{2}^{n}$. I want to check whether $X$ is smooth or not. 
In the case that you have a $n\in\mathbb{Z}_{>1}$ and an algebraically closed field $k$ such that $\text{char}(k)$ does not divide $n$ and $n-1$, one can easily check that $X$ is indeed smooth. Here you can use the fact that $X = (X\cap D(x_{0}))\cup (X\cap D(x_{2}))$, where $D(x_{i}) = \{(x_{0}:x_{1}:x_{2})\in\mathbb{P}^{2}\rvert x_{i}\neq 0\}$. One can use the isomorphisms
$\varphi_{0}: X\cap D(x_{0})\rightarrow Z(x_{0,1}^{n}-x_{0,2}+x_{0,2}^{n}), (x_{0}:x_{1}:x_{2})\mapsto (\frac{x_{1}}{x_{0}},\frac{x_{2}}{x_{0}})$ and 
$\varphi_{2}: X\cap D(x_{2})\rightarrow Z(x_{2,1}^{n}-x_{2,0}^{n-1}+1), (x_{0}:x_{1}:x_{2})\mapsto (\frac{x_{0}}{x_{2}},\frac{x_{1}}{x_{2}})$,
where $x_{i,j} = \frac{x_{j}}{x_{i}}$. 
For points $P\in X$ with $x_{2}=0$ we know that $x_{1}=0$ and consequently $x_{0} \neq 0$. Then we consider $\varphi_{0}$ and consequently we have the matrix $\begin{pmatrix} 0 & -1\end{pmatrix}$, so $X$ is smooth at $P$. 
For points $P\in X$ with $x_{2}\neq 0$ we can consider $\varphi_{2}$ and consequently we have the matrix \begin{pmatrix} -(n-1)x_{2,0}^{n-2} & nx_{2,1}^{n-1}\end{pmatrix}. By the assumption that $\text{char}(k)$ does not divide $n$ and $n-1$ one can find that the matrix is not equal to the zero matrix, so $X$ is smooth at $P$. 
As soon as you have a $n\in\mathbb{Z}_{>1}$ and an algebraically closed field $k$ such that $\text{char}(k)$ divides $n$ or $n-1$ the problems arrive. I also know that if $P\in X$ and $P$ lies on two irreducible components of $X$, then $P$ is singular. 
My Question/Problem: How does one check if $X$ is smooth or not in the case that we have a $n\in\mathbb{Z}_{>1}$ and an algebraically closed field $k$ such that $\text{char}(k)$ divides $n$ or $n-1$? 
Definition of Smoothness: Let $X$ be a variety. For $P\in X$, $X$ is smooth at $P$ if there is an open subvariety $U$ of $X$ containing $P$ and an isomorphism $\varphi:U\rightarrow Z(f_{1},...,f_{n-d})\subset\mathbb{A}^{n}$ for some $d\leqslant n$ and $f_{1},...,f_{n-d}$, such that the rank of the $n-d$ by $n$ matrix over $k$:
$\begin{pmatrix} \frac{\partial f_{i}}{\partial x_{j}}(\varphi(P)) \end{pmatrix}_{i,j}$ equals $n-d$, i.e. is maximal. 
 A: You've done a very good job so far and are just a hair away from a complete solution. The only thing you need to do is use the fact that if $\operatorname{char} k$ divides $n$, then $n=0$ in $k$ (and similarly if $\operatorname{char} k$ divides $n-1$, then $n-1=0$ inside $k$). Once you do this and rewrite the Jacobian matrix after making these simplifications, you'll exactly see what you need to finish. I've provided full details in some spoiler tags below for you to check your work once you've done this.

 First, we note that we can upgrade our Jacobian to work directly with the homogeneous polynomial to avoid the need to split in to cases, as is proven here. Then our homogeneous Jacobian is given by the matrix $$J=\begin{pmatrix} (n-1)x_0^{n-2}x_2 & -nx_1^{n-1} & -nx_2^{n-1} \end{pmatrix}.$$

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 If $\operatorname{char} k$ divides $n$, then $n=0$ inside $k$, so $J = \begin{pmatrix} -x_0^{n-2}x_2 & 0 & 0\end{pmatrix}$ which has rank $0$ iff $x_0=0$ or $x_2=0$. We may verify that if $x_2=0$, this gives that $X$ is singular at $[1:0:0]$ and if $x_0=0$, at all points $[0:a:b]$ so that $a^n+b^n=0$. So $X$ is singular in this case.

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 If $\operatorname{char} k$ divides $n-1$, then $n-1=0$ inside $k$, so $J=\begin{pmatrix} 0 & -x_1^{n-1} & -x_2^{n-1}\end{pmatrix}$ which has rank $0$ iff $x_0=x_2=0$. There are no points with $x_0=x_2=0$ on $X$, though, so $X$ is smooth in this case.

Edit: In the comments, the asker revealed that their primary concern was with whether $X$ smooth at $P$ implies $\varphi_0(X\cap D(x_0))$ smooth at $\varphi_0(p)$. This is true: The intersection of $U$ and $X\cap D(x_0)$ may be covered by principal open affine subsets of the form $U_f=(X\cap D(x_0))_g$ for $f$ a function on $U$ and $g$ a function on $X\cap D(x_0)$, so there's one of these that contains $x$. If $\varphi(U)\cong V(f_1,\cdots,f_n)\subset \Bbb A^n$ then $\varphi(U)_f\cong V(f_1,\cdots,f_n,fx_{n+1}−1)\subset \Bbb A^{n+1}$ and we can see that the Jacobian matrix here is full rank on one side iff it's full rank on the other. Apply this again for $X\cap D(x_0)$ to get the result. (Put more succicntly, smoothness is a local condition.)
