Given expansion is $\left(\frac{x^2}{a}-\frac{b}{x}\right)^{11}$.
$\therefore$ The general term is
$\begin{aligned} T_{r+1} & { }^{11} C_r\left(\frac{x^2}{a}\right)^{11-r}\left(-\frac{b}{x}\right)^r \\ & ={ }^{11} C_r(x)^{22-3 r}(-b)^r\left(\frac{1}{a}\right)^{11-r}\end{aligned}$
$\therefore$ For coefficient $x^7$, put $22-3 r=7$
$\begin{aligned} \Rightarrow & & 3 r & =15 \\ \Rightarrow & & r & =5\end{aligned}$
and for coefficient of $x^4$, put $22-3 r=4$
$\begin{aligned} & \Rightarrow \quad 3 r=18 \Rightarrow r=6 \\ & \therefore \quad T_6={ }^{11} C_5\left(\frac{1}{a}\right)^6(-b)^5 \\ & \text { and } T_7={ }^{11} C_6\left(\frac{1}{a}\right)^6(-b)^6\end{aligned}$
According to the given condition,
$\begin{array}{rlrl} & { }^{T_6}+T_7=0 \\ & { }^{11} C_5\left(\frac{1}{a}\right)^6(-b)^5+{ }^{11} C_6\left(\frac{1}{a}\right)^5(-b)^6=0 \\ \Rightarrow & { }^{11} C_5\left(\frac{1}{a}\right)^5(-b)^5\left(\frac{1}{a}-b\right)=0 \\ \Rightarrow & & \frac{1}{a}-b=0 \\ \Rightarrow & a b=1\end{array}$