Answer
See below
Work Step by Step
Let $S$ be the subspace of $R^3$ spanned by the vectors $f_1(x) =e^x \\ f_2(x)=e^{-x}\\f_3(x) = \sinh(x)$.
We can see that $\frac{1}{2}f_1(x)\\=\frac{1}{2}e^x-\frac{1}{2}e^{-x}\\=\frac{e^x-e^{-x}}{2}\\
=f_3(x)$
Then we have $f(x)=af_1(x)+bf_2(x)+cf_3(x)\\
=af_1(x)+bf_2(x)+c(\frac{1}{2}f_1(x)-\frac{1}{2}f_2(x))\\
=af_1(x)+bf_2(x)+\frac{1}{2}cf_1(x)-\frac{1}{2}cf_2(x)\\
=(a+\frac{1}{2}c)f_1(x)+(b-\frac{1}{2}c)f_2(x)\\
\rightarrow f \in span \{f_1,f_2\}\\
\rightarrow span\{f_1,f_2,f_3\} \subset span \{f_1,f_2\}$
Since $\{f_1,f_2\} \subset \{f_1,f_2,f_3\}\\
\rightarrow span\{f_1,f_2\} \subset \{f_1,f_2,f_3\}\\
\rightarrow S=span\{f_1,f_2\}$
Let $a,b$ be scalars such that
$ae^x+be^{-x}=0 \forall x \in (-\infty,\infty)\\
\rightarrow ae^x-be^{-x}=0 \forall x \in(-\infty, \infty)$
Substitute $x=0 \rightarrow a+b=0, a-b=0$
then $a+b+a-b=0\\
2a=0\\
a=0\\
\rightarrow b=0$
Hence, the linear combination of $f_1$ and $f_2$ is trivial then $f_1,f_2$ are linearly independent in $(-\infty, \infty)$
Hence, $\{f_1,f_2\}$ is a basic for $S \rightarrow \dim[S]=2 $
