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Let L be modular lattice, M be lattice and $f:Lto M$ be a homomorphism. I want to show $f(L)$ is a modular lattice..

We already know that homomorphic image of lattice is lattice.
So we only want to show that if $f(a)leq f(b)$ then $f(a) vee (f(x)wedge f(b))= (f(a)vee f(x))wedge f(b) $ for $a,b,x in L$

Since L is modular $aleq b$ implies $a vee(xwedge b)= (avee x)wedge b$

My problem is:

I must begin with the assumption $f(a)leq f(b)$ then show $f(a) vee (f(x)wedge f(b))= (f(a)vee f(x))wedge f(b) $ for which I need to use $a vee(xwedge b)= (avee x)wedge b$

But $f(a)leq f(b)$ need not necessarily imply $aleq b$ since it is only homomorphism and not isomorphism.

A lattice is modular if and only if it satisfies the modular law:
$$ale bimplies alor(xland b)=(alor x)land b.tag1$$
Equivalently, a lattice is modular if and only if it satisfies the modular identity:
$$(cland b)lor(xland b)=[(cland b)lor x)land b.tag2$$
Although $(1)$ and $(2)$ and not equivalent in isolation, they are equivalent in the presence of the other axioms of lattice theory; namely, their equivalence follows from the fact that, in a lattice, $ale b$ if and only if $a=cland b$ for some $c$.

Now, since $(2)$ is an identity, it is preserved by homomorphisms; therefore, a homomorphic image of a modular lattice is a modular lattice.

P.S. Here is a straightforward presentation which avoids mentioning the "modular identity":

Given $a,b,xin L$ with $f(a)le f(b)$, we want to show that
$$f(a)lor(f(x)land f(b))=(f(a)lor f(x))land f(b).tag3$$
We don't know if $ale b$, so let $a_0=aland b$. Then $a_0le b$, so by the modular law we have
$$a_0lor(xland b)=(a_0lor x)land b.tag4$$
Moreover, $f(a_0)=f(aland b)=f(a)land f(b)=f(b)$ since $f(a)le f(b)$. Since $f$ is a homomorphism, and since $f(a_0)=f(a)$, we see that $(3)$ follows from $(4)$.

You can instead replace $b$ with $avee b$, since $f(avee b) =f(a) vee f(b) =f(b) $.