3. As shown in the figure, points $M, N$ are on the sides $A B, A C$ of $\triangle A B C$, respectively, and $\overrightarrow{A M}=$ $x \overrightarrow{A B}, \overrightarrow{A N}=y \overrightarrow{A C}, D$ is the midpoint of segment $B C$, and $G$ is the intersection of segment $M N$ and $A D$. If $\overrightarrow{A G}=\frac{3}{5} \overrightarrow{A D}$, then the minimum value of $\frac{1}{x^{2}}+\frac{4}{y^{2}}$ is $\qquad$
$$
\overrightarrow{A G}=\frac{3}{5} \overrightarrow{A D}=\frac{3}{10}(\overrightarrow{A B}+\overrightarrow{A C})=\frac{3}{10 x} \overrightarrow{A M}+\frac{3}{10 y} \overrightarrow{A N}
$$

From $M, G, N$ being collinear, we get $\frac{3}{10 x}+\frac{3}{10 y}=1 \Rightarrow \frac{1}{x}+\frac{1}{y}=\frac{10}{3} \Rightarrow \frac{1}{x}=\frac{10}{3}-\frac{1}{y}$. $\frac{1}{x^{2}}+\frac{4}{y^{2}}=\left(\frac{10}{3}-\frac{1}{y}\right)^{2}+\frac{4}{y^{2}}=\frac{5}{y^{2}}-\frac{20}{3} \cdot \frac{1}{y}+\frac{100}{9}=5\left(\frac{1}{y}-\frac{2}{3}\right)^{2}+\frac{80}{9}$, given $\frac{1}{y} \geqslant 1, \frac{1}{x}=\frac{10}{3}-\frac{1}{y} \geqslant 1 \Rightarrow \frac{1}{y} \in\left[1, \frac{7}{3}\right]$,
so when $\frac{1}{y}=1$, the minimum value of $\frac{1}{x^{2}}+\frac{4}{y^{2}}$ is $5-\frac{20}{3}+\frac{100}{9}=\frac{85}{9}$.