derivative of inverse matrix by itself Let $A$ be a matrix, supposedly $k\times k$ matrix. 
I know that 
$$\frac{\partial A^{-1}}{\partial A} = -A^{-2} $$
I do not know how I am supposed to obtain the following results using this fact. I want to know the step of
$$\frac{\partial a^\top A^{-1} b}{\partial A} = -(A^\top)^{-1}ab^\top (A^\top)^{-1} $$
Also, I want to know the solution to
$$\frac{\partial (A^\top)^{-1}ab^\top (A^\top)^{-1} }{\partial A} = ? $$

Start with the defining equation for the matrix inverse and find its differential.
$$\eqalign{
 I &= A^{-1}A \\
 0 &= dA^{-1}\,A + A^{-1}\,dA \\
 dA^{-1} &=  -A^{-1}\,dA\,A^{-1} \\
}$$
Next note the gradient of a matrix with respect to itself.
$$
{\mathcal H}_{ijkl}
 = \frac{\partial A_{ij}}{\partial A_{kl}}
 = \delta_{ik}\delta_{jl}
$$
Note that ${\mathcal H}$ is a 4th order tensor with some interesting symmetry properties (isotropic). It is also the identity element for the Frobenius product, i.e. for any matrix $B$
$${\mathcal H}:B=B:{\mathcal H}=B$$
Now we can answer your first question. The  function of interest is scalar-valued. Let's find its differential and gradient
$$\eqalign{
 \phi &= a^TA^{-1}b \cr &= ab^T:A^{-1} \\
d\phi &= ab^T:dA^{-1} \cr &= -ab^T:A^{-1}\,dA\,A^{-1}  \\
      &= -A^{-T}ab^TA^{-T}:dA \\
\frac{\partial\phi}{\partial A} &= -A^{-T}ab^TA^{-T} \\
}$$
Now let's try the second question. This time the function of interest is matrix-valued.
$$\eqalign{
 F &= A^{-1}ab^TA^{-1} \\
dF &= dA^{-1}ab^TA^{-1} + A^{-1}ab^TdA^{-1} \\
  &= -A^{-1}\,dA\,A^{-1}ab^TA^{-1} - A^{-1}ab^TA^{-1}\,dA\,A^{-1} \\
  &= -A^{-1}\,dA\,F - F\,dA\,A^{-1} \\
  &= -\Big(A^{-1}{\mathcal H}F^T + F{\mathcal H}A^{-T}\Big):dA \\
\frac{\partial F}{\partial A}
  &= -\Big(A^{-1}{\mathcal H}F^T+F{\mathcal H}A^{-T}\Big) \\
}$$
This gradient is a 4th order tensor.
If you prefer, you can vectorize the matrices to flatten the result.
$$\eqalign{
{\rm vec}(dF) &= -{\rm vec}(A^{-1}\,dA\,F + F\,dA\,A^{-1}) \\
  &= -(F^T\otimes A^{-1} + A^{-T}\otimes F)\,{\rm vec}(dA) \\
df &= -(F^T\otimes A^{-1} + A^{-T}\otimes F)\,da \\
\frac{\partial f}{\partial a}
  &= -\Big(F^T\otimes A^{-1} + A^{-T}\otimes F\Big) \\\\
}$$
In some step above, a colon was used to denote the Frobenius (double-contraction) product
$$\eqalign{
A &= {\mathcal H}:B &\implies &A_{ij}
  &= \sum_{kl}{\mathcal H}_{ijkl} B_{kl} \\
\alpha &= H:B &\implies &\alpha
  &= \sum_{ij}H_{ij} B_{ij} = {\rm Tr}(H^TB) \\
}$$
