I was looking at the following circuit. AFAIK, this is a low-pass Sallen–Key filter, with a VCVS instead of the op-amp.
I started to work out the transfer function, but somehow I got lost at a certain step.
First, you can use the common node to state:
$$V_{A}(s)=V_{1}(s)$$and$$V_{D}(s)=\mu V_{C}(s)=V_{2}(s)$$then the KCL for nodes at \$V_b\$ and \$V_C(s)\$:
$$\frac{V_{B}(s)-V_{1}(s)}{R_{1}}+\frac{V_{B}(s)-V_{C}(s)}{R_{2}}+\frac{V_{B}(s)-\mu V_{C}(s)}{\frac{1}{C_{1}s}}=0$$$$\frac{V_{C}(s)-V_{B}(s)}{R_{2}}+\frac{V_{C}(s)}{\frac{1}{C_{2}s}}=0$$
multiplicating by \$R_{1}R_{2}\$$$\frac{R_{1}R_{2}(V_{B}(s)-V_{1}(s))}{R_{1}}+\frac{R_{1}R_{2}(V_{B}(s)-V_{C}(s))}{R_{2}}+\frac{R_{1}R_{2}(V_{B}(s)-\mu V_{C}(s))}{\frac{1}{C_{1}s}}=0$$
$$\frac{R_{1}R_{2}(V_{C}(s)-V_{B}(s))}{R_{2}}+\frac{R_{1}R_{2}(V_{C}(s))}{\frac{1}{C_{2}s}}=0$$
$$R_{2}(V_{B}(s)-V_{1}(s))+R_{1}(V_{B}(s)-V_{C}(s))+\frac{R_{1}R_{2}(V_{B}(s)-\mu V_{C}(s))}{\frac{1}{C_{1}s}}=0$$
$$R_{1}(V_{C}(s)-V_{B}(s))+\frac{R_{1}R_{2}(V_{C}(s))}{\frac{1}{C_{2}s}}=0$$
$$R_{2}V_{B}(s)-R_{2}V_{1}(s)+R_{1}V_{B}(s)-R_{1}V_{C}(s)+\frac{R_{1}R_{2}V_{B}(s)}{\frac{1}{C_{1}s}}-\frac{R_{1}R_{2}\mu V_{C}(s)}{\frac{1}{C_{1}s}}=0$$
$$R_{1}V_{C}(s)-R_{1}V_{B}(s)+\frac{R_{1}R_{2}V_{C}(s)}{\frac{1}{C_{2}s}}=0$$
$$R_{2}V_{B}(s)-R_{2}V_{1}(s)+R_{1}V_{B}(s)-R_{1}V_{C}(s)+C_{1}sR_{1}R_{2}V_{B}(s)-C_{1}sR_{1}R_{2}\mu V_{C}(s)=0$$
$$R_{1}V_{C}(s)-R_{1}V_{B}(s)+C_{2}sR_{1}R_{2}V_{C}(s)=0$$
$$(R_{1}+R_{2}+R_{1}R_{2}C_{1}s)V_{B}(s)-(R_{1}+\mu R_{1}R_{2}C_{1}s)V_{C}(s)=R_{2}V_{1}(s)$$
$$(R_{1}+R_{1}R_{2}C_{2}s)V_{C}(s)=R_{1}V_{B}(s)$$
$$(R_{1}+R_{2}+R_{1}R_{2}C_{1}s)V_{B}(s)-(R_{1}+\mu R_{1}R_{2}C_{1}s)V_{C}(s)=R_{2}V_{1}(s)$$
$$R_{1}(1+R_{2}C_{2}s)V_{C}(s)=R_{1}V_{B}(s)$$$$\frac{R_{1}(1+R_{2}C_{2}s)V_{C}(s)}{R_{1}}=\frac{R_{1}V_{B}(s)}{R_{1}}$$
$$(1+R_{2}C_{2}s)V_{C}(s)=V_{B}(s)$$
From the schematic you can tell
$$V_{2}(s)=\mu V_{C}(s)$$
$$(R_{1}+R_{2}+R_{1}R_{2}C_{1}s)(1+R_{2}C_{2}s)V_{C}(s)-(R_{1}+\mu R_{1}R_{2}C_{1}s)V_{C}(s)=R_{2}V_{1}(s)$$
$$(R_{1}+R_{2}+R_{1}R_{2}C_{1}s)(1+R_{2}C_{2}s)V_{C}(s)-(R_{1}V_{C}(s))-R_{1}R_{2}C_{1}s\mu V_{C}(s)=R_{2}V_{1}(s)$$
$$(R_{1}+R_{2}+R_{1}R_{2}C_{1}s)(1+R_{2}C_{2}s)=R_{1}+R_{1}R_{2}C_{2}s+R_{2}+R_{2}^{2}C_{2}s+R_{1}R_{2}C_{1}s+R_{1}R_{2}^{2}C_{1}C_{2}s^{2}$$
$$(R_{1}+R_{1}R_{2}C_{2}s+R_{2}+R_{2}^{2}C_{2}s+R_{1}R_{2}C_{1}s+R_{1}R_{2}^{2}C_{1}C_{2}s^{2}-R_{1}-R_{1}R_{2}C_{1}s\mu)V_{C}(s)=R_{2}V_{1}(s)$$
$$(R_{1}R_{2}C_{2}s+R_{2}+R_{2}^{2}C_{2}s+R_{1}R_{2}C_{1}s+R_{1}R_{2}^{2}C_{1}C_{2}s^{2}-R_{1}R_{2}C_{1}s\mu)V_{C}(s)=R_{2}V_{1}(s)$$
But here, I understand the way to go is using the fact of \$V_{2}(s)=\mu V_{C}(s)\$, but I got lost, how to group it to get the \$H(s)\$?