Step 1:

It is given that $xy$ is the point of contact of the curve and its tangent.

Hence the slope=$\large\frac{ y+3}{x+4}$.Let this be $m_1$

but we know that slope of the tangent to the curve is $\large\frac{dy}{dx}$.Let this be $m_2$

According to the given information, $m_2= 2m_1$

$\large\frac{dy}{dx}=\frac{ 2(y+3)}{x+4}$

Seperating the variables

we get $\large\frac{dy}{(y+3)} =\frac{ 2dx}{(x+4)}$

Step 2:

Integrating on both sides

we get $\int\large\frac{ dy}{(y+3)}=$$2\int\large\frac{dy}{x+4}$

or $\log (y+3)=2\log (x+4)+\log c$

$\log (y+3)=\log c . (x+4)^2$

$y+3=c(x+4)^2$

Step 3:

It is given that the curve passes through the point $-2,1$

Substituting for $x$ and $y$ in the general equation to evaluate for $c$ we get

$1+3=c(-2+4)^2$

$4=4c$

$c=1$

substituting this for $c$

we get $y+3=(x+4)^2$

This is the required equation of the curve