I have a bijective continuous function $f$, which maps an $(n\times1)$ dimensional column vector $t=[t_1,...t_n]'$ to another $(n\times1)$ dimensional column vector $f(t)=[f_1(t),...f_n(t)]'$. I define the matrix derivative $\frac{df(t)}{dt}$ as follows:
$$\frac{df(t)}{dt}=\begin{bmatrix}\frac{\partial f_1(t)}{\partial t_1} & ... & \frac{\partial f_n(t)}{\partial t_1} \\ \vdots & \ddots & \vdots \\ \frac{\partial f_1(t)}{\partial t_n} & ... & \frac{\partial f_n(t)}{\partial t_n} \end{bmatrix}$$
Let $a$ be a fixed vector of dimension $(n \times 1)$. I want to find a $t_0$ satisfying the below, expressed in terms of $A$ and $f$ (and the derivatives of $f$ if necessary):
$$\frac{d f(t)}{d t}\bigg|_{t=t_0}a-f(t_0)=0$$
Where the $|_{t=t_0}$ represents that the derivative is being evaluated at $t=t_0$. Is it possible to find an analytic solution to this problem? If so, how? And where can I read more on this sort of problem? Perhaps there is a tensor based solution?
This isn't even feasible when $n = 1$. Let $f$ be any horrible transcendental expression or polynomial with roots that cannot be expressed in radicals you like, say $$ f(t) = \begin{pmatrix} t_1^5 - t_1 - 1 \end{pmatrix} \text{.} $$ Then your equation is $$ (5 t_0^4 - 1)a - (t_0^5 - t_0 - 1) = 0 \text{,} $$ the same thing as $$ -t_0^5 +5at_0^4 +t_0-a+1 = 0 \text{,} $$ which, for instance, does not have roots representable by radicals for any $a$ which is an integer in $[-10,10]$ except $a = 0$. So in the lowest dimensional case, there is no analytical formula that works for general $a$. The best one can say is "whatever the roots of this polynomial are".
An example in the transcendental direction is to take $f(t) = \sin \cos t_1$. Then you want to solve $$ -a (\cos \cos t_0)\sin t_0 - \sin \cos t_0 = 0 \text{,} $$ which is hopeless.
Without serious constraints on the functions appearing in $f$, the problem you pose contains the general version of "find the zeroes of an arbitrary function", which is hopeless.