Considering only the principal values of the inverse trigonometric functions, the value off

$\tan <!--\; \; -->({\sin }^{-<!--\; -\; -->1}<!--\; \; -->\left(\frac{3}{5}\right)-<!--\; -\; -->2{\cos }^{-<!--\; -\; -->1}<!--\; \; -->\left(\frac{2}{\sqrt{5}}\right))$

We want to evaluate the given expression:
$\tan \left({\sin }^{-1}\right(\frac{3}{5})-2{\cos }^{-1}(\frac{2}{\sqrt{5}}\left)\right)$

Let us simplify the terms inside the tangent function individually.

First, let $x={\sin }^{-1}\left(\frac{3}{5}\right)$.
This means:
$\sin \left(x\right)=\frac{3}{5}$
Since $\frac{3}{5}>0$, the principal value of $x$ lies in the first quadrant, i.e., $x\in (0,\frac{\pi }{2})$.
Using the identity $\cos \left(x\right)=\sqrt{1-{\sin }^2\left(x\right)}$:
$\cos \left(x\right)=\sqrt{1-{\left(\frac{3}{5}\right)}^2}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$
Therefore, we have:
$\tan \left(x\right)=\frac{\sin \left(x\right)}{\cos \left(x\right)}=\frac{3/5}{4/5}=\frac{3}{4}$
Which gives:
$x={\tan }^{-1}\left(\frac{3}{4}\right)$

Next, let $y={\cos }^{-1}\left(\frac{2}{\sqrt{5}}\right)$.
This means:
$\cos \left(y\right)=\frac{2}{\sqrt{5}}$
Since $\frac{2}{\sqrt{5}}>0$, the principal value of $y$ also lies in the first quadrant, i.e., $y\in (0,\frac{\pi }{2})$.
Using the identity $\sin \left(y\right)=\sqrt{1-{\cos }^2\left(y\right)}$:
$\sin \left(y\right)=\sqrt{1-{\left(\frac{2}{\sqrt{5}}\right)}^2}=\sqrt{1-\frac{4}{5}}=\sqrt{\frac{1}{5}}=\frac{1}{\sqrt{5}}$
Therefore, we have:
$\tan \left(y\right)=\frac{\sin \left(y\right)}{\cos \left(y\right)}=\frac{1/\sqrt{5}}{2/\sqrt{5}}=\frac{1}{2}$
Which gives:
$y={\tan }^{-1}\left(\frac{1}{2}\right)$

Now we substitute these values back into the expression, which becomes $\tan (x-2y)$.
First, let us calculate $2y=2{\tan }^{-1}\left(\frac{1}{2}\right)$ using the double-angle formula $2{\tan }^{-1}\left(u\right)={\tan }^{-1}\left(\frac{2u}{1-u^2}\right)$ for $\left|u\right|<1$:
$2{\tan }^{-1}\left(\frac{1}{2}\right)={\tan }^{-1}\left(\frac{2·(1/2)}{1-{(1/2)}^2}\right)={\tan }^{-1}\left(\frac{1}{1-1/4}\right)={\tan }^{-1}\left(\frac{1}{3/4}\right)={\tan }^{-1}\left(\frac{4}{3}\right)$

Now, we can find the value of $\tan (x-2y)$:
$\tan (x-2y)=\tan \left({\tan }^{-1}\right(\frac{3}{4})-{\tan }^{-1}(\frac{4}{3}\left)\right)$
Applying the subtraction identity for tangent, $\tan (A-B)=\frac{\tan \left(A\right)-\tan \left(B\right)}{1+\tan \left(A\right)\tan \left(B\right)}$:
$\tan (x-2y)=\frac{\frac{3}{4}-\frac{4}{3}}{1+\left(\frac{3}{4}\right)\left(\frac{4}{3}\right)}$
Simplifying the numerator and denominator:
$\text{Numerator}=\frac{3}{4}-\frac{4}{3}=\frac{9-16}{12}=\frac{-7}{12}$
$\text{Denominator}=1+1=2$
Thus, the final value is:
$\tan (x-2y)=\frac{-7/12}{2}=\frac{-7}{24}$