How to do this using @RenderBody from Asp.net MVC?

A solution to use Ajax , without having to change all% from Views to PartialViews , would be to Ajax request on Action and then only get RenderBody generated from View .

An example to get clearer:

_Layout

(...)
<div id="Conteudo">
    @RenderBody()
</div>
(...)

Then on your View :

<a onclick="Open('Modulo/Index')">Clique aqui</a>
<div id="Conteudo2">
</div>

<script>
    function Open(url) {
        url = '@Url.Content("~/")' + url;
        $.ajax({
            url: url,
            type: 'GET',
            success: function (response) {
                $('#Conteudo2').html($(response).find('#Conteudo'));
            },
            error: function () {
                alert('Ocorreu um erro!');
            }
        });
    }
</script>

I find AJAX uninteresting for this solution.

It seems to me that the content of this side menu is directly related to the access link in the top menu. You can pass this list of side menu links dynamically (by ViewBag or something like that) and render it normally in _Layout.

On _Layout:

<div class="conteudo">
    <div class="menulateral">
        @foreach(var item in ViewBag.LinksFromMenuSuperior) {
            <a href="@item.Link">@item.Titulo</a>
        }
    </div>
    <div>
        @RenderBody()
    </div>
</div>

You can also use @Section {} and @RenderSection (), which you may not know because you did not comment as an option, but worth a query, unfortunately it (in this solution) would behave much like @PartialView () because you will also have to change the pages.