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Let $V(\vec{r})=-k\frac{m}{r}$, $\vec{r}(t)=(x(t), y(t))$, $\dot{\vec{r}}=(v_x(t),v_y(t))$ and put the sun at the origin. Using Newtons second law show that the following equations hold- $\dot{x}=v_x,$ $\dot{y}=v_y,$ $\dot{v}_x=-k\frac{x}{(x^2+y^2)^{\frac{3}{2}}},$ $\dot{v}_y=-k\frac{y}{(x^2+y^2)^{\frac{3}{2}}}.$

Edit: After some help I have arrived at ${\vec{a}}=\dot{\vec{v}}=-k\frac{(x,y)}{(x^2+y^2)^\frac{3}{2}}.$ Is this correct and if so how do I proceed from here?

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    Look up a decent classical mechanics text. I used Marion many years ago. It takes several pages to get there. – 2017-02-19
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    What is the actual question? Your title says "Deriving Kepler's laws" and your question say "show that the following equations hold". Which is it? – 2017-02-19
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    How do I derive with respect to time? Are the coordinates not function of $\theta$? I assume we are using polar coordinates. – 2017-02-19
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    Sorry Spencer. The problem is part of a larger assignment to do with Kepler's laws. – 2017-02-19
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    I see, so you merely want to know how to derive the indicated equations. – 2017-02-19
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    You really should rephrase the title of this question by the way. Something like "how to determine the acceleration due to a spherically symmetric potential" would be appropriate. – 2017-02-19

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First of all $v_x$ and $v_y$ are equal to $\dot{x}$ and $\dot{y}$ by definition.

Now Newton's second law relates the force on a particle to its acceleration,

$$ \vec{F} = m \dot{\vec{v}},$$

From this it is easy to see that $\dot{v}_x=F_x/m=-\frac1m \frac{\partial V}{\partial x}$ and $\dot{v}_y=F_y/m=-\frac1m \frac{\partial V}{\partial y}$. The partial derivatives are easy to compute, just remember that $r=\sqrt{x^2+y^2}$.

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    Ok so I have gotten $\vec{a}=\dot{\vec{v}}=-k\frac{(x,y)}{(x^2+y^2)^\frac{3}{2}}$. Which seems correct. What is the next step? Trying to take the partial derivative of the denominater (x,y) is throwing me off. – 2017-02-19
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    Anybody please? I have being working on the problem the past 3 hours and am not making any headway. Any further help greatly appreciated. – 2017-02-19
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    @Lucon, you now have an answer to the first step of your problem. Deriving the rest of Kepler's laws is really outside the scope of this question. If you want help deriving a particular law at this point, then asking a new question would be more effective. – 2017-02-19
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    You may find these resources helpful : http://astro.physics.uiowa.edu/~srs/2961_12/Lec08_notes.pdf http://astro.physics.uiowa.edu/~srs/2961_12/Lec09_notes.pdf – 2017-02-19