Use implicit differentiation with respect to #x# to get: #12x-2y-2x\cdot dy/dx+9y^2\cdot dy/dx=0# (note that the Chain Rule is used). Now solve for #dy/dx# to get #dy/dx=(2y-12x)/(-2x+9y^2)#. Finally, plug in the point #(x,y)=(6,0)# to get #dy/dx|_{(x,y)=(6,0)}=\frac{-72}{-12}=6#.
By the way, it is worth confirming that #(x,y)=(6,0)# is on the original curve: #6\cdot 6^{2}-2\cdot 6\cdot 0+3\cdot 0^{3}=216#.
Also, we are assuming #y# is a function of #x# when we differentiate the original equation. In fact, you may find it more helpful to rewrite the original equation as #6x^2-2xf(x)+3(f(x))^3=216# and then, after differentiation, write #12x-2f(x)-2xf'(x)+9(f(x))^2f'(x)=0#. Now solve for #f'(x)# to get #f'(x)=(2f(x)-12x)/(-2x+9(f(x))^2)# and plug in #x=6# and #f(6)=0# as above.
Here's a picture of the situation in this problem. The tangent line at the given point is shown as well (dashed).
