Since $x$ is not an independent variable, but a function of $t$, we have the following situation: We are given the function
$$F(t,u):=f\bigl(x(t)-(k+u)t\bigr)-u=0\ ,$$
assumed of class $C^1$. The equation
$$F(u,t)=0$$ then implicitly defines $u$ as a function of $t$ under favorable circumstances. The implicit function theorem has the following to say in this regard: Given a point $(t_0,u_0)$ satisfying $F(t_0,u_0)=0$ there is a $C^1$-function $$\psi: \>t\mapsto u:=\psi(t)\qquad(t_0-h<t<t_0+h),\qquad \psi(t_0)=u_0\ ,$$ such that
$$F\bigl(t,\psi(t)\bigr)\equiv0\qquad(t_0-h<t<t_0+h)\ .$$
Furthermore one has the formula
$$u'(t_0)=-{F_t(t_0,u_0)\over F_u(t_0,u_0)}={f'\bigl(x(t_0)-(k+u_0)t_0\bigr)\bigl(x'(t_0)-(k+u_0)\bigr)\over f'\bigl(x(t_0)-(k+u_0)t_0\bigr)t_0+1}\ .\tag{1}$$
The all important assumption for all this to hold is that $F_u(t_0,u_0)\ne0$, i.e., that the denominator in $(1)$ does not vanish.