Tangent at #x=x_0# for a curve given by #y=f(x)# has slope of #f'(x_0)# where #f'(x)=(df)/(dx)#. Also tangent passes through #(x_0,f(x_0)#. Hence, we use these two to get the equation of tangent in point slope form.
Here our curve is #f(x)=1/x^3-1/(x-3)^3#
and we are seeking tangent at #x=1/4# and #f(1/4)=1/(1/4)^3-1/(1/4-3)^3=64-(-64/1331)=64 64/1331#i.e. #(1/4,64 64/1331)# or #(1/4,85248/1331)#
and #f'(x)=-3/x^4-(-3/(x-3)^4)=-3/x^4+3/(x-3)^4#
and #f'(1/4)=-3/(1/4)^4+3/(1/4-3)^4=-768+3xx256/14641#
= #-768+768/14641=-11243520/146641#
As slope is #-11243520/146641# and line passes through #(1/4,85248/1331)#, its equation is
#y-85248/1331=-11243520/146641(x-1/4)#
and multiplying by #146641# we get equation of tangent is
#14641y-85248xx11=-11243520x+2810880#
or #11243520x+14641y-3748608=0#
The tangent appears as shown (not drawn to scale and shrunk vertically).
graph{(1/x^3-1/(x-3)^3-y)(11243520x+14641y-3748608)=0 [-1, 1, 0, 150]}
