Consider an algebraic curve $X$ defined by an equation $y^2 = f(x)$ where $f$ is a polynomial of degree $d \geq 3$ with coefficients in a field $k$.
For $m = 2$ and $d = 3$, we can get an equation of the form $f(x) = x^3 + Ax + B$ (if the characteristic of $k$ is not $2$ or $3$). Then for this case, a general result says that $$ \Delta \neq 0 \Longleftrightarrow X \text{ is non-singular (i.e. an elliptic curve)}. $$
Question: Is there something similar to a discriminant of $X$ for polynomials $f$ of arbitrary degrees? (i.e. $X$ is a superelliptic curve)
If $k$ is any field and $F(x,y):=y^n-f(x)$ with $n\geq 2$ an integer and $deg(f)=d$, you may consider the algebraic curve $C:=Spec(A)$ where $A:=k[x,y]/(F(x,y))$. The jacobian ideal $J$ given by
$$J:=(ny^{n-1},-f'(x))$$
defines the singular subscheme $C_{sing}:=V(J) \subset C$. Assume for simplicity $char(k) \neq n$. Let $(a,b)\in C_{sing}(K)$ be a $K$ rational point of $C_{sing}$, where $k\subseteq K$ is a field. It follows $a,b\in K$ and $f(a)=f'(a)=0$. Hence the curve $C$ has a $K$-rational singular point $(a,0)\in K^2$ iff
$$f(a)=f'(a)=0.$$
If $f(x):=a_0+a_1x+\cdots +a_dx^d$ we define the discriminant $\Delta(f)$ to be the resultant $\Delta(f):=Res(f(x),f'(x))$ of the polynomial $f(x)$ and it's derivative $f'(x)$. It follows $\Delta(f)=0$ iff $f(x)$ has a multiple root in $K$, where $K$ is a finite extension of $k$. Hence the discriminant $\Delta(f)=0$ iff the curve $C$ has a $K$-rational singular point, where $k\subseteq K$ is a finite extension.
Question: "Is there something similar to a discriminant of X for polynomials f of arbitrary degrees? (i.e. $X$ is a superelliptic curve)?"
Answer: Yes, it seems to me you may use the classical discriminant $\Delta(f)$ of the polynomial $f(x)$ as explained above. If $char(k) \neq n$ it follows $\Delta(f)=0$ iff $C_{sing}(K) \neq \emptyset$ for some finite extension $k\subseteq K$.
Note: The discriminant $\Delta(f):=Res(f(x),f'(x))$ is a polynomial in the coefficients of $f(x)$, hence you can calculate $\Delta(f)$ without knowing anything on the roots of $f(x)$ or if the roots of $f(x)$ live in a finite extension $k \subseteq K$. There is a well known determinantal formula for the resultant $Res(f(x),g(x))$ of any pair of polynomials $f(x),g(x)$ - the "Sylvester determinant".
As remarked in the comments: If $k$ is not algebraically closed you may get singular points defined over non-trivial extensions $k \subseteq K$. The curve $C$ is finite type over $k$ and closed points $x\in C$ have residue field $k \subseteq \kappa(x)$ which is a finite extension in general. Hence points of $C_{sing}$ are not in general defined over the base field $k$.
Example: If $k=\mathbb{Q}$ is the field of rational numbers and $f(t):=g(t)^2$ where $g(t):=t^2-2$. Let $F(x,y):=y^n-f(t)$. It follows
$$f'(t)=2g(t)g'(t)=4t(t^2-2) \in k[t].$$
Hence $f(t)=0$ and $f'(t)=0$ does not have a solution in $k$. It has a solution in $k(\sqrt{2})$. This phenomenon also happens in characteristic $p>0$.
Note: If the polynomial $f(x)\in k[x^p]$ is a polynomial in $x^p$ it follows $f'(x)=0$ hence $\Delta(f)=Res(f(x), f'(x))$ is identically zero in characteristic $p>0$. Over the algebraic closure $\overline{k}$ it follows $f(x^p)=g(x)^p$, hence $f(x)$ has multiple roots and the curve is singular.