Derivatives

Substitute $x=2$ in the equation 
\[x^3+y^3+244=x^2f(x)    (1)\]
to get 
\[2^3+y^3+244=2^2f(2)=2^2(9)=36.\]
Hence
\[y^3=36-244-8=-216=-6^3 \Rightarrow y=-6.\]
So the line passes through the point (i) $(2,-6)$.

To find the slope we differentiate (1) to get
\[3x^2+3y^2 y'=2xf(x)+x^2f'(x).\]
Substitute $x=2$ and $y=-6$ to get 
\[3(2)^2+3(-6)^2 y'=2(2)f(2)+2^2f'(2)=4\times 9+4\times 0=36.\]
So 
\[12+3\times 36 y'=36 \Rightarrow y'=\frac{24}{3\times 36}=\frac{8}{36}=\frac{2}{9}.\]

The slope of the line is (ii) $\frac{2}{9}$.

The equation of the tangent line is 
\[y-(-6)=\frac{2}{9}(x-2)\]
or 
\[y=\frac{2}{9}x-\frac{4}{9}-6=\frac{2}{9}x-\frac{58}{9}.\]
Hence equation of the tangent line is
(iii) \[y=\frac{2}{9}x-\frac{58}{9}.\]