PHP / CSS onmouseover和onmouse out不适用于PHP echo

Here I want to style my table with css. But when I try to add this events inside echo, it says,

Parse error: syntax error, unexpected 'this' (T_STRING), expecting ',' or ';' in E:\xampp\htdocs\Ceylon Aviation\Flights.php on line 146

As I'm a beginner to PHP, I can't figure out the exact error. Any helpful suggestions to edit my code? Thank You!

My Code:

echo "<tr onmouseover = /"this.style.backgroundColor = /'#ffff66' ; " onmouseout = /"this.style.backgroundColor = /'#d4e3e5';">";

This works too:

    echo "<tr onmouseover='this.style.backgroundColor=#ffff66' onmouseout='this.style.backgroundColor=#d4e3e5'>";

Change it to call javascript functions: in changeColor function

function changeColor(){ this.style.backgroundColor = '#ffff66'; }

Point one, onmouseover and onmouseout are JavaScript events.
Point two, you can use the :hover from css.
Point three, you only need to echo the html

var div = document.getElementById("js");
//set width and height
   div.style.width = "100px";
   div.style.height = "100px";
//set the start color
   div.style.backgroundColor = "#F0F";

function OnMouseOver(o) {
   o.style.backgroundColor = "#F00";
}

function OnMouseOut(o) {
   o.style.backgroundColor = "#F0F";
}

#css {
  width: 100px;
  height: 100px;
  background-color: #0FF;
  
}

#css:hover {
  background-color: #00F;
}

<div id="js" onmouseover="OnMouseOver(this)" onmouseout="OnMouseOut(this)">
</div>
<div id="css">
</div>

I added a snippet to show both css and JavaScript ways

</div>