Finding the equation of a tangent line that passes through a curve at point P. I'm not sure how to long the question in the title, I have a curve $y = lnx^{3}$ and the point $P=(0,1)$. I have the derivative of the function $y' = \frac{3}{x}$, but I'm not sure where to go from here. Can anyone point me in the right direction?
Let's find the general equation of the tangent to this curve.
Let's write $ln(x^3)$ as $3ln(x)$.

You know the slope. You know one point through which it passes,say,$(h,3ln(h))$.
You are done. Equation of line:

$y-3ln(h)=\frac 3 h (x-h)$   

It passes through $(0,1)$

Lets try plugging that in $(x.y)$ in line equation.

$1-3ln(h)=\frac 3 h (0-h)$

As h is not zero as function is not defined there, we can cancel it.

$1-3ln(h)=-3$

$ln(h)=4/3$

$h=e^{\frac4 3}$

Lets plug it in equation of line

$y-3 \frac 4 3 =\frac {3} {h}(x-h)$

$y-4=\frac {3x} h - 3 $

$y-1=\frac {3x}{e^{\frac 4 3 }}$

Congos...You did it.
Let's find the general equation of the tangent to this curve.
Let's write $ln(x^3)$ as $3ln(x)$.

You know the slope. You know one point through which it passes,say,$(h,3ln(h))$.
You are done. Equation of line:

$y-3ln(h)=\frac 3 h (x-h)$   

It passes through $(0,1)$

Lets try plugging that in $(x.y)$ in line equation.

$1-3ln(h)=\frac 3 h (0-h)$

As h is not zero as function is not defined there, we can cancel it.

$1-3ln(h)=-3$

$ln(h)=4/3$

$h=e^{\frac4 3}$

Lets plug it in equation of line

$y-3 \frac 4 3 =\frac {3} {h}(x-h)$

$y-4=\frac {3x} h - 3 $

$y-1=\frac {3x}{e^{\frac 4 3 }}$

Congos...You did it.