1. Interpret the expression

   The given expression is $\cot\left(\cosec^{-1}\frac{5}{3}+\tan^{-1}\frac{2}{3}\right).$

   Let $A=\cosec^{-1}\frac{5}{3}, \qquad B=\tan^{-1}\frac{2}{3}.$

   We need to find $\cot(A+B).$

2. Find trigonometric values for $A$

   Since $\csc A=\frac{5}{3},$ we get $\sin A=\frac{3}{5}.$

   Taking $A$ in the principal range of $\cosec^{-1}$, $A$ is acute here, so $\cos A=\frac{4}{5}.$

   Therefore, $\tan A=\frac{\sin A}{\cos A}=\frac{3/5}{4/5}=\frac{3}{4}.$

3. Find trigonometric values for $B$

   Since $B=\tan^{-1}\frac{2}{3},$ we have $\tan B=\frac{2}{3}.$

4. Use the tangent addition formula

   $\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}.$

   Substituting, $\tan(A+B)=\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\cdot\frac{2}{3}}.$

   First compute the numerator: $\frac{3}{4}+\frac{2}{3}=\frac{9+8}{12}=\frac{17}{12}.$

   Now the denominator: $1-\frac{3}{4}\cdot\frac{2}{3}=1-\frac{1}{2}=\frac{1}{2}.$

   Hence, $\tan(A+B)=\frac{17/12}{1/2}=\frac{17}{12}\cdot 2=\frac{17}{6}.$

5. Find the cotangent

   $\cot(A+B)=\frac{1}{\tan(A+B)}=\frac{1}{17/6}=\frac{6}{17}.$

6. Check options

   $\frac{6}{17}$ matches Option A.

Final Answer: $\boxed{\frac{6}{17}}$