# How do you find the slope of the tangent line to the curve 6x^2 - 2xy + 3y^3 = 216 at the point 6,0?

Use implicit differentiation with respect to $x$ to get: $12 x - 2 y - 2 x \setminus \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + 9 {y}^{2} \setminus \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 0$ (note that the Chain Rule is used). Now solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$ to get $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 y - 12 x}{- 2 x + 9 {y}^{2}}$. Finally, plug in the point $\left(x , y\right) = \left(6 , 0\right)$ to get $\frac{\mathrm{dy}}{\mathrm{dx}} {|}_{\left(x , y\right) = \left(6 , 0\right)} = \setminus \frac{- 72}{- 12} = 6$.
By the way, it is worth confirming that $\left(x , y\right) = \left(6 , 0\right)$ is on the original curve: $6 \setminus \cdot {6}^{2} - 2 \setminus \cdot 6 \setminus \cdot 0 + 3 \setminus \cdot {0}^{3} = 216$.
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 $6 {x}^{2} - 2 x f \left(x\right) + 3 {\left(f \left(x\right)\right)}^{3} = 216$ and then, after differentiation, write $12 x - 2 f \left(x\right) - 2 x f ' \left(x\right) + 9 {\left(f \left(x\right)\right)}^{2} f ' \left(x\right) = 0$. Now solve for $f ' \left(x\right)$ to get $f ' \left(x\right) = \frac{2 f \left(x\right) - 12 x}{- 2 x + 9 {\left(f \left(x\right)\right)}^{2}}$ and plug in $x = 6$ and $f \left(6\right) = 0$ as above. 