Which Schrödinger equation is correct? In the coordinate representation, in 1D, the wave function depends on space and time, $\Psi(x,t)$, accordingly the time dependent Schrödinger equation is
$$H\Psi(x,t) = i\hbar\frac{\partial}{\partial t}\Psi(x,t).\tag{1}$$
In a representation-free notations we deal instead with the ket $|\Psi\rangle = |\Psi(t)\rangle$. Now how to write Schrödinger equation in this case?
I find some books write
$$H|\Psi\rangle = i\hbar\frac{\partial}{\partial t}|\Psi\rangle,\tag{2}$$
and others write
$$H|\Psi\rangle = i\hbar\frac{d}{dt}|\Psi\rangle.\tag{3}$$
So which one of the last 2 equations is correct?
 A: I) Let us reformulate OP's question as:

What time differentiation symbol should one use on the right-hand side of the time-dependent ket Schrödinger equation?

Answer:

Whatever symbol that means
$$  \lim_{\Delta t \to 0} \frac{|\Psi(t+\Delta t )\rangle_S-|\Psi(t) \rangle_S }{\Delta t}.$$

So apparently both OP's suggestions work. Here the subscript "$S$" (and "$H$") denotes the Schrödinger (Heisenberg) picture, where bras and kets evolve (are unchanged) and operators are unchanged (evolve), respectively.
II) Let us mention for completeness that in the Heisenberg picture:

*

*$$|\Psi \rangle_H\quad \text{does not evolve in time}.$$


*$$ {}_H\langle x,t |\quad \text{does not evolve in time}.$$


*$$ \psi (x,t) ~=~ {}_H\langle x,t |\Psi \rangle_H.$$


*$$\begin{align} {}_H\langle x,t |\hat{x}(t)~=~& x ~{}_H\langle x,t |  
\cr~\Updownarrow~&\cr
{}_H\langle x,t |\hat{x}(t)|\Psi \rangle_H ~=~& x \psi (x,t). \end{align}$$


*$$\begin{align}{}_H\langle x,t |\hat{p}(t) ~=~& \frac{\hbar}{i} \frac{\partial}{\partial x} ~{}_H\langle x,t |, 
\cr~\Updownarrow~&\cr
{}_H\langle x,t |\hat{p}(t)|\Psi \rangle_H ~=~& \frac{\hbar}{i} \frac{\partial}{\partial x} \psi (x,t).\end{align} $$


*$$\begin{align} i\hbar\frac{\partial}{\partial t}  {}_H\langle x,t | ~=~&{}_H\langle x,t |\hat{H}(t) 
\cr~\Updownarrow~&\cr
i\hbar\frac{\partial}{\partial t}\psi (x,t) ~=~&{}_H\langle x,t |\hat{H}(t) |\Psi \rangle_H\cr
~=~&(-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}+V(x))\psi (x,t) .\end{align}$$
For a full explanation, see e.g. J.J. Sakurai, Modern Quantum Mechanics.
A: All your equations are correct. 
$$H\Psi(x,t) = i\hbar\frac{\partial}{\partial t}\Psi(x,t)$$
says that we will differentiate wrt to time, keeping x constant. 
$$H|\Psi\rangle = i\hbar\frac{d}{dt}|\Psi\rangle$$
says that we differentiate wrt to time, and moreover this is all the state vector $|\Psi\rangle$ depends. 
$$H|\Psi\rangle = i\hbar\frac{\partial}{\partial t}|\Psi\rangle$$
has the same content as the second equation, its just that we don't need to be careful by writing a partial derivative since there isn't any other variables in $|\Psi\rangle$ besides time anyway, so either a partial or total derivative will do. 
So in all 3 equations we are differentiating wrt to the same time dependance, its just that sometimes we have to be careful if there are other variables around or not. 
A: Your second and third equations are the same equation. They just use a different notation for the time derivative. Since in this "abstract" form $|\Psi \rangle$ only depends on time perhaps it is more correct to use the last one, but it is matter of taste.
In order to get your first equation (a wave equation), you must project on $\langle x|$:
$$H(P=-i\hbar \partial _x , X=x)\Psi (x,t)=i\hbar \partial _t \Psi (x,t)$$
In this case $\partial _t$ (rather than $\frac{d}{dt}$) is a better notation because now $\Psi$ also depends on the coordinate $x$.
Regarding Nick Kidman's comment:
i) In the abstract SE the Hamiltonian $H$ is a function of "abstract" operators $H=H(P, X)$ (capital letters refer to operators). 
ii) One has the canonical commutation relations $[X,P]=i\hbar$. A realization or representation of this commutation relation in a (certain) space of functions $f(x)$ (the lower case $x$ is a coordinate instead an operator) $X\,f(x)=x\,f(x)$ and $Pf(x)=-i\hbar\partial _x \,f(x)$ since $[x,-i\hbar\partial _x]f(x)=i\hbar f(x)$. (One can prove that this is the only representation of the commutation relation modulo unitarity equivalence, in a finite-dimensional system. In QFT one has truly different representations.) Therefore $P|x\rangle =-i\hbar\partial _x \,|x\rangle$ and $X|x\rangle =x \,|x\rangle$
iii) By definition $\Psi (x)\equiv \langle x|\Psi \rangle$.
A: Just remind yourself what the curly d means. It is just denoting a partial derivative so it is 1D and the function is only dependent on t then it is:
$$\frac{df(t)}{dt}$$ 
In all cases where the function is dependent on more than one variable (e.g. x and t) then you must use the curly d to denote that you are doing a partial derivative. 
