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

Let $3x+y=t$

$3+\frac{dy}{dx}=\frac{dt}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{dt}{dx}-3$

Putting $\frac{dy}{dx}$in differential equation

$\frac{dt}{dx}-3={\left(t+4\right)}^2$

$\Rightarrow \frac{dt}{dx}={\left(t+4\right)}^2+3$

$\Rightarrow \frac{dt}{{\left(t+4\right)}^2+3}=\frac{dx}{1}$

Integrating both sides

$\Rightarrow \int \frac{dt}{{\left(t+4\right)}^2+3}=\int \frac{dx}{1}$

As we know,$\int \frac{dt}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+k$

$\Rightarrow \frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{t+4}{\sqrt{3}}\right)=x+k$

Replacing value of $t$

$\frac{1}{\sqrt{3}}\left({\tan }^{-1}\left(\frac{3x+y+4}{\sqrt{3}}\right)\right)-x=k$

Compare with

$\frac{1}{\sqrt{3}}\left({\tan }^{-1}\left(f\left(x,y\right)\right)\right)-x=k$

$f\left(x,y\right)=\frac{3x+y+4}{\sqrt{3}}$

$f(1,2)=\frac{3+2+4}{\sqrt{3}}=3\sqrt{3}$