\(x=\frac{7+4y}{3}\Rightarrow3x^2+4y^2=3.\left(\frac{7+4y}{3}\right)^2+4y^2=\frac{\left(7+4y\right)^2}{3}+4y^2\)
\(=\frac{49+56y+16y^2+12y^2}{3}=\frac{49+56y+28y^2}{3}\)
\(=\frac{28.\left(\frac{7}{4}+2y+y^2\right)}{3}=\frac{28.\left(y^2+2y+1+\frac{3}{4}\right)}{3}=\frac{28\left(y+1\right)^2+21}{3}\)
\(\ge\frac{21}{3}=7\)