From the Maclaurin series expansion, one has
$$
\ln (1+x)= -\sum_{n=1}^\infty\frac{(-1)^n}nx^n, \quad |x|<1, \tag1
$$ one deduces
$$
\begin{align}
\ln[(1+3x)^{(1+x)}]&=(1+x)\ln(1+3x)
\\\\&=-(1+x)\sum_{n=1}^\infty\frac{(-1)^n}n3^nx^n
\\\\&=-\sum_{n=1}^\infty\frac{(-1)^n}n3^nx^n-\sum_{n=1}^\infty\frac{(-1)^n}n3^nx^{n+1}
\\\\&=3x-\sum_{n=2}^\infty\frac{(-1)^n}n3^nx^n-\sum_{n=2}^\infty\frac{(-1)^{n-1}}{n-1}3^{n-1}x^{n}
\\\\&=3x+\sum_{n=2}^\infty\frac{(-1)^{n-1}(2n-3)}{n(n-1)}3^{n-1}x^n
\\\\&=3x+\sum_{n=1}^\infty\frac{(-1)^n(2n-1)}{n(n+1)}3^{n}x^{n+1}. \tag2
\end{align}
$$
By putting $x=-\dfrac19$ in $(2)$, one gets
$$
\sum_{n=1}^{\infty} \frac{1-2n}{3^n n(n+1)}=3-8\ln\frac 32. \tag3
$$
This will be rather difficult.
Let's first of all suppose there are no problems, i.e. the functions you are interested in are infinitely differentiable and continuous in the range of values you are interested in.
Then, even with the simple example that you know the Maclaurin series of $f(x)$ and want the Taylor series of $f(x)$ about another point $a$, this is a cumbersome endeavour. Here is the procedure.
With the known Maclaurin series (you memorized the $m_k$)
$f(x) = \sum\limits_{k = 0}^{\infty} \frac{f^{(k)} (0)}{k!} x^k = \sum\limits_{k = 0}^{\infty} m_k x^k$
you require the Taylor series
$f_a(x) = \sum\limits_{k = 0}^{\infty} c_k (x-a)^k$
Now the coefficients are given by
$c_k = \frac{f^{(k)} (a)}{k!}$
which in turn you can get by differentiating your MacLaurin series and evaluate at $a$:
$c_k = \frac{f^{(k)} (a)}{k!} = \sum\limits_{n = k}^{\infty} \frac{n!}{k! (n-k)!} m_n a^{n-k}$
E.g.
$c_0 = \sum\limits_{n = 0}^{\infty} m_n a^{n}$
So you need to sum an infinite series to arrive at your new coefficients - big labour.
Another method is to identify the coefficients of all powers of $(x-a)$. To do so, write the MacLaurin series
$f(x) = \sum\limits_{k = 0}^{\infty} m_k ((x-a)+a)^k
= \sum\limits_{k = 0}^{\infty} m_k \sum_{j=0}^k {k \choose j} (x-a)^j (a)^{k-j}$
This leads to the same result.
Things get worse if you consider $g(f(x))$.
I guess this already answers your question: it's not a smooth technique to calculate the Taylor expansion from the known MacLaurin series.
Best Answer
$$ e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+....\\turn\\x\\to \\-x\\so\\e^{-x}=1-x+\frac{(-x)^2}{2!}+\frac{(-x)^3}{3!}+\frac{(-x)^4}{4!}+....\\so\\e^{-0.9}$$