Point of intersection of pair of straight lines. Let non homogenous curve equation be $ax^2 + 2hxy + by^2 + 2gx + 2fy + c=0 $ then first time partial differentiating with respect to x gives $ ax + hy + g =0$ and second time partial differentiating with respect to y gives $by + hx + f =0$ solving both equations give point of intersection of pair of lines. $$\left(\dfrac{hf-bg}{ab - h^2}\right )  , \left (  \dfrac {gh-af}{ab - h^2}\right)$$
My question is - why partial differentiating makes work easy and what those two equations represent?
 A: You have some errors in your partial derivatives:
We start with your original equation: $$ax^2 + 2hxy + by^2 + 2gx + 2fy + c=0 $$
$(1)$ Finding the partial derivative with respect to $x$ gives: $$ax +hy +g=0$$  
$(2)$ Finding the partial derivative  with respect to $y$ gives $$2hx+2by+2f = by +hx+ f = 0$$  So in the first case, w.r.t. x, your $b$ in $by$ needs to be an $h$.  In the second partial, w.r.t. y,  your $us$ should be $hx.$
That gives us the equations of two lines: 
$$\begin{align} ax + hy+g &= 0 \\ \\
hx+by +f &= 0\end{align}$$
Now we have a system of two equations and two variables, with $a, b, c, f, g, h$ all real numbers, from which we seek the solution $(x, y)$ at which the two given lines intersect.
Indeed, sollving the system gives us the intersection of the partial derivatives at $$\left(\dfrac{hf-bg}{ab - h^2}\right )  , \left (  \dfrac {gh-af}{ab - h^2}\right)$$ provided $ab-h^2 \leq 0$ and $b\neq 0$.
A: Suppose the lines are $L_1: ax+by+c=0$ and $L_2: dx+ey+f=0$. Then equation of pair of straight lines is $L_1 L_2 =0$. Now on partial differentiation w.r.t. $x$ and $y$ we get the following
$$aL_2+dL_1=0,\,\, \text{ and } bL_2+eL_1=0.$$
Now using concept of linear algebra we know that these two equations have same solution as the original equations $L_1=0$ and $L_2=0$. (we can verify it easily). Hence solving the equations obtained on partial differentiation will give the point of intersection of two lines.
Hope that answers your question well.
