What you have listed is a state equation. Usually a state-space representation has two equations:
(i) $\dot{x}(t)=Ax(t)+Bu(t)$
(ii) $y(t)=Cx(t)+Du(t)$
One takes the input $u(t)$ and generates the state $x(t)$ and another that takes the input $u(t)$ and state $x(t)$ and generates the output $y(t)$. You are missing the second equation.
That aside, the equation that you have is basically a 1st order linear differential equation. It's important to note, here, that the input is not $x(t)$, it is $u(t)$. $x(t)$ is the state variable.
I will examine just the first half (input to state) and leave the second half (input and state to output) to you. I will remark, though, that if $x(t)$ is linear with respect to $u(t)$ then $y$ in equation (ii) above necessarily has to be linear by virtual of adding two linear terms together, so the second half is really trivial.
The first half is:
(1) $\dot{x}(t)=Ax(t)+Bu(t)$
Can you show that it is linear?
I will show additivity and leave homogeneity to you. Say you have inputs $u_1$, $u_2$, and $u_3=u_1+u_2$.
One way to do this is to solve (1). As it is a 1st order ODE, it is not too difficult, but let's be lazy.
Let $x_1$ be the state generated by input $u_1$ and $x_2$ be the state generated from input $u_2$, then we have
(2) $\dot{x_1}(t)=Ax_1(t)+Bu_1(t)$
and
(3) $\dot{x_2}(t)=Ax_2(t)+Bu_2(t)$
adding (2) and (3) gives
(4) $\dot{x_1}(t)+\dot{x_2}(t)=Ax_1(t)+Ax_2(t)+Bu_1(t)+Bu_2(t)$
From linearity of derivative, we get that:
(5) $\dot{x_3}(t)=A(x_3(t))+B(u_3(t))$
where $x_3 = x_1+x_2$.
So input $u_3 = u_1+u_2$ generates state variable $x_3 = x_1+x_2$, showing that additivity holds for the state equation.